🧠:[<think>Okay, so I need to find the smallest positive root of the equation 2 sin(6x) + 9 cos(6x) = 6 sin(2x) + 7 cos(2x). Hmm, let's see. I remember that when dealing with equations involving multiple trigonometric functions, sometimes it helps to use identities to simplify them or rewrite them in a form that's easier to solve. First, let me look at both sides of the equation. On the left side, we have 2 sin(6x) + 9 cos(6x), and on the right side, 6 sin(2x) + 7 cos(2x). The arguments of the trigonometric functions are 6x and 2x. Maybe I can express both sides in terms of the same multiple angle? Or perhaps use a substitution to make the equation simpler. Let me think.Alternatively, I remember that expressions like A sin θ + B cos θ can be written as C sin(θ + φ) or C cos(θ + φ), where C is the amplitude and φ is the phase shift. Maybe rewriting both sides in that form would help me compare them more easily. Let me try that.Starting with the left side: 2 sin(6x) + 9 cos(6x). The general formula is A sin θ + B cos θ = C sin(θ + φ), where C = √(A² + B²) and tan φ = B/A. So here, A = 2 and B = 9. Therefore, C = √(2² + 9²) = √(4 + 81) = √85. Then φ = arctan(B/A) = arctan(9/2). So the left side can be written as √85 sin(6x + arctan(9/2)). Similarly, the right side: 6 sin(2x) + 7 cos(2x). Here, A = 6 and B = 7, so C = √(6² + 7²) = √(36 + 49) = √85. Interesting, same amplitude! Then φ = arctan(7/6). Therefore, the right side can be written as √85 sin(2x + arctan(7/6)).So the equation simplifies to:√85 sin(6x + arctan(9/2)) = √85 sin(2x + arctan(7/6))Since √85 is positive and non-zero, we can divide both sides by √85 to get:sin(6x + arctan(9/2)) = sin(2x + arctan(7/6))Now, when does sin α = sin β? That occurs when α = β + 2πn or α = π - β + 2πn for some integer n. So we can set up two equations:1. 6x + arctan(9/2) = 2x + arctan(7/6) + 2πn2. 6x + arctan(9/2) = π - (2x + arctan(7/6)) + 2πnLet's simplify both equations.Starting with the first equation:6x + arctan(9/2) = 2x + arctan(7/6) + 2πnSubtract 2x from both sides:4x + arctan(9/2) = arctan(7/6) + 2πnThen subtract arctan(9/2):4x = arctan(7/6) - arctan(9/2) + 2πnSo,x = [arctan(7/6) - arctan(9/2) + 2πn] / 4Similarly, for the second equation:6x + arctan(9/2) = π - 2x - arctan(7/6) + 2πnBring all x terms to the left and constants to the right:6x + 2x = π - arctan(7/6) - arctan(9/2) + 2πn8x = π - arctan(7/6) - arctan(9/2) + 2πnTherefore,x = [π - arctan(7/6) - arctan(9/2) + 2πn] / 8Now, we need to find the smallest positive x from these solutions. So we need to evaluate these expressions for different integer values of n and see which gives the smallest positive x.First, let's compute arctan(7/6) and arctan(9/2). Let me approximate these values.arctan(7/6): 7/6 is approximately 1.1667. The arctangent of 1 is π/4 ≈ 0.7854, and arctan(√3) ≈ 1.0472 (π/3). arctan(1.1667) is a bit more than π/4. Let's calculate it using a calculator. Hmm, arctan(7/6) ≈ arctan(1.1667) ≈ 0.86217 radians.arctan(9/2): 9/2 is 4.5. arctan(4.5) is a large angle. Since tan(1.35 radians) ≈ tan(77.3 degrees) ≈ 4.3315, tan(1.4) ≈ tan(80.2 degrees) ≈ 5.671. So 4.5 is between tan(1.35) and tan(1.4). Let's compute arctan(4.5). Maybe using a calculator again: arctan(4.5) ≈ 1.3526 radians.So arctan(7/6) ≈ 0.86217 and arctan(9/2) ≈ 1.3526.Let's compute the first case:x = [0.86217 - 1.3526 + 2πn] / 4 ≈ [ -0.4904 + 6.2832n ] / 4We need x positive, so -0.4904 + 6.2832n > 0 => 6.2832n > 0.4904 => n >= 1.For n=0: negative, so discard.For n=1:x ≈ ( -0.4904 + 6.2832 ) / 4 ≈ 5.7928 / 4 ≈ 1.4482 radians.Wait, but x is in terms of radians? Wait, no. Wait, the variable x is just a variable; the equation is in terms of x, but when we solve for x, the solution is in radians as the argument of sine and cosine. Wait, actually, no. Wait, in trigonometric equations, unless specified, the solutions are in radians. So x is in radians. So 1.4482 radians is approximately 83 degrees.But let's check the second case.Second case:x = [π - 0.86217 - 1.3526 + 2πn] / 8 ≈ [3.1416 - 0.86217 - 1.3526 + 6.2832n] / 8 ≈ [0.9268 + 6.2832n] / 8So for n=0:x ≈ 0.9268 / 8 ≈ 0.11585 radians ≈ 6.64 degrees.For n=1: x ≈ (0.9268 + 6.2832)/8 ≈ 7.21/8 ≈ 0.901 radians ≈ 51.6 degrees.So between the two cases, the first case gives x ≈1.4482 for n=1, and the second case gives x ≈0.11585 for n=0, which is much smaller. So perhaps the smallest positive root is approximately 0.11585 radians. But let's verify if this is correct.However, before trusting these approximations, maybe we can compute the exact expressions or check if there are smaller solutions.Wait, but perhaps the first case with n=1 gives x≈1.448, while the second case with n=0 gives x≈0.1158. So 0.1158 is smaller. But we need to confirm if this is indeed a solution.Alternatively, maybe there's a solution between 0 and 0.1158? Let me check by testing the original equation at x=0 and near x=0.At x=0: Left side: 2 sin(0) +9 cos(0)=9. Right side:6 sin(0)+7 cos(0)=7. So 9 vs 7; left side is greater. Then as x increases from 0, let's see when the two sides cross.Wait, maybe plotting or testing the function f(x) = 2 sin(6x) +9 cos(6x) -6 sin(2x) -7 cos(2x) would help. We can check the value at x=0: f(0)=9-7=2. At x=0.1 radians:Compute f(0.1):Left side: 2 sin(0.6) +9 cos(0.6). sin(0.6)≈0.5646, cos(0.6)≈0.8253. So 2*0.5646≈1.1292, 9*0.8253≈7.4277. Total left≈8.5569.Right side:6 sin(0.2)+7 cos(0.2). sin(0.2)≈0.1987, cos(0.2)≈0.9801. So 6*0.1987≈1.1922, 7*0.9801≈6.8607. Total right≈8.0529.So f(0.1)=8.5569 -8.0529≈0.504. Still positive.At x=0.1158 radians (approx solution from second case):Compute f(0.1158):Left: 2 sin(6*0.1158) +9 cos(6*0.1158). 6*0.1158≈0.6948. sin(0.6948)≈0.6411, cos(0.6948)≈0.7675. So 2*0.6411≈1.2822, 9*0.7675≈6.9075. Total left≈8.1897.Right:6 sin(2*0.1158) +7 cos(2*0.1158). 2*0.1158≈0.2316. sin(0.2316)≈0.2293, cos≈0.9733. 6*0.2293≈1.3758, 7*0.9733≈6.8131. Total right≈8.1889.So f(x)=8.1897 -8.1889≈0.0008. Almost zero. So that's very close. So x≈0.1158 is indeed a root. But is there a smaller positive root? Let's check between x=0 and x=0.1158.Wait, at x=0, f(x)=2. At x=0.1158, f(x)=0.0008. So it's decreasing from 2 to almost 0. So it's possible that the first crossing is at x≈0.1158. But let me check at x=0.05:Left: 2 sin(0.3) +9 cos(0.3). sin(0.3)≈0.2955, cos≈0.9553. So 2*0.2955≈0.591, 9*0.9553≈8.598. Total≈9.189.Right:6 sin(0.1)+7 cos(0.1). sin(0.1)≈0.0998, cos≈0.9950. 6*0.0998≈0.5988, 7*0.9950≈6.965. Total≈7.5638. So f(0.05)=9.189 -7.5638≈1.625. Still positive.At x=0.1, f(x)=0.504. At x=0.1158, f(x)=0.0008. So it's decreasing from x=0 to x=0.1158. So maybe the first root is around 0.1158. But wait, the function is decreasing here. Wait, but f(x) is positive at x=0.1158, just barely. Wait, but earlier calculation showed f(0.1158)=0.0008, which is still positive. So maybe it crosses zero just slightly above 0.1158.Wait, perhaps my approximation is slightly off. Let me compute more accurately.Let me use more precise values.First, let's compute arctan(7/6) and arctan(9/2). Using a calculator:arctan(7/6) ≈ 0.8621700546 radians.arctan(9/2) ≈ 1.352627526 radians.So the expression for x in the second case with n=0 is:x = [π - arctan(7/6) - arctan(9/2)] / 8Plug in the numbers:π ≈ 3.14159265363.1415926536 - 0.8621700546 - 1.352627526 ≈ 3.1415926536 - 2.2147975806 ≈ 0.926795073 radiansThen x ≈ 0.926795073 / 8 ≈ 0.115849384 radians.So x≈0.115849384. Let's compute f(x) at this x.First, compute 6x: 6 * 0.115849384 ≈ 0.6950963 radians.sin(0.6950963) ≈ sin(0.6951) ≈ 0.64114583, cos(0.6951) ≈ 0.7674559So left side: 2*0.64114583 + 9*0.7674559 ≈ 1.28229166 + 6.9071031 ≈ 8.18939476.Right side: compute 2x=0.231698768 radians.sin(0.231698768) ≈ 0.2297525, cos≈0.97326156*0.2297525 ≈ 1.378515, 7*0.9732615 ≈ 6.8128305. Total ≈1.378515 +6.8128305≈8.1913455.So f(x)=8.18939476 -8.1913455≈-0.0019507. Wait, that's negative. But when I computed with approximate x=0.1158, f(x)≈0.0008. Hmm, perhaps there's a discrepancy here. Let me check the calculations again.Wait, maybe my calculator is more precise. Wait, let's recompute with exact x.Given x ≈0.115849384 radians.Compute left side: 2 sin(6x) +9 cos(6x). 6x=0.695096304.sin(0.695096304):Using Taylor series around 0.7 radians:sin(0.7) ≈ 0.644217687.But 0.695096304 is 0.7 - 0.004903696.sin(a - b) ≈ sin a cos b - cos a sin b.Compute sin(0.7 - 0.004903696):≈ sin(0.7)cos(0.004903696) - cos(0.7)sin(0.004903696)cos(0.004903696)≈1 - (0.004903696)^2/2 ≈1 - 0.000012 ≈0.999988sin(0.004903696)≈0.004903696 - (0.004903696)^3/6≈0.004903696 - 0.000000196≈0.0049035So ≈0.644217687 *0.999988 - 0.764842187 *0.0049035≈0.644217687*0.999988≈0.644210.764842187*0.0049035≈0.003747So sin(0.695096304)≈0.64421 -0.003747≈0.64046Similarly, cos(0.695096304):cos(a - b)=cos a cos b + sin a sin b≈cos(0.7)*0.999988 + sin(0.7)*0.0049035cos(0.7)≈0.764842187sin(0.7)≈0.644217687So ≈0.764842187*0.999988 +0.644217687*0.0049035≈0.76483 +0.003156≈0.767986Therefore, left side: 2*0.64046 +9*0.767986≈1.28092 +6.91187≈8.19279Right side:6 sin(2x) +7 cos(2x). 2x=0.231698768 radians.sin(0.231698768):Using Taylor series for small angles:sin(0.2317)≈0.2317 - (0.2317)^3/6≈0.2317 - 0.00215≈0.22955cos(0.2317)≈1 - (0.2317)^2/2≈1 - 0.0267≈0.9733So 6*0.22955≈1.3773, 7*0.9733≈6.8131. Total≈1.3773 +6.8131≈8.1904Therefore, f(x)=8.19279 -8.1904≈0.00239. So positive. Wait, but in my previous calculation with x=0.115849384, using approximate sin and cos, got left≈8.19279, right≈8.1904, so f(x)=0.00239. So positive. But earlier when I computed with more precise sin and cos, I thought it was negative. Hmm. Wait, perhaps the exact value crosses zero near this x. Therefore, the approximate solution x≈0.1158 is close to the actual root, but may not be exact. Therefore, perhaps we need to use a numerical method to find a better approximation.Alternatively, since we derived this solution from the equation sin(6x + arctan(9/2)) = sin(2x + arctan(7/6)), leading to x≈0.1158, but the exact solution requires solving 8x = π - arctan(7/6) - arctan(9/2). But maybe we can compute arctan(7/6) + arctan(9/2) exactly?Wait, there's a formula for arctan a + arctan b = arctan((a + b)/(1 - ab)), but that's only valid under certain conditions. Let me check.Actually, arctan a + arctan b = arctan[(a + b)/(1 - ab)] + kπ, depending on the quadrant. Let's see.Let me compute (7/6 + 9/2)/(1 - (7/6)(9/2)) = (7/6 + 27/6)/(1 - 63/12) = (34/6)/(-51/12) = (17/3)/(-17/4) = (17/3)*(-4/17) = -4/3.So arctan(7/6) + arctan(9/2) = arctan(-4/3) + kπ. Since arctan(7/6) is positive and arctan(9/2) is positive, their sum is positive. arctan(-4/3) is negative, so we need to add π to get into the correct quadrant. Therefore,arctan(7/6) + arctan(9/2) = π + arctan(-4/3) = π - arctan(4/3)Therefore, π - arctan(7/6) - arctan(9/2) = π - [π - arctan(4/3)] = arctan(4/3)Wait, wait, let me redo that step.If arctan(7/6) + arctan(9/2) = π - arctan(4/3). Let me verify:Let’s set θ = arctan(7/6) + arctan(9/2). Then tan θ = tan(arctan(7/6) + arctan(9/2)) = [7/6 + 9/2]/[1 - (7/6)(9/2)] = [ (7/6 + 27/6) ] / [1 - 63/12] = [34/6]/[ -51/12 ] = (17/3)/(-17/4) = -4/3. Therefore, tan θ = -4/3. Since both arctan(7/6) and arctan(9/2) are in the first quadrant, their sum θ is in the second quadrant (since tan θ is negative). Therefore, θ = π - arctan(4/3). Therefore,π - arctan(7/6) - arctan(9/2) = π - θ = π - (π - arctan(4/3)) = arctan(4/3)Therefore, the expression for x in the second case becomes:x = [arctan(4/3)] / 8Therefore, x = (arctan(4/3)) / 8So instead of approximating arctan(7/6) and arctan(9/2), we can use arctan(4/3). arctan(4/3) is a known angle. Let's compute it.arctan(4/3) is approximately 0.927295218 radians. Therefore, x = 0.927295218 / 8 ≈ 0.1159119 radians. So this is a more precise value. So x≈0.1159119 radians.Therefore, this is the exact solution from the second case with n=0. So this should be the smallest positive root. Let's check this value.x≈0.1159119 radians.Compute 6x≈0.6954715 radians.sin(0.6954715)≈sin(0.6955)= let's use calculator: sin(0.6954715)=approx 0.6413, cos(0.6954715)=approx 0.7673.Left side: 2*0.6413 +9*0.7673 ≈1.2826 +6.9057≈8.1883.Right side:2x≈0.2318238 radians.sin(0.2318238)=approx0.2299, cos=approx0.9732.6*0.2299≈1.3794, 7*0.9732≈6.8124. Total≈1.3794 +6.8124≈8.1918.So f(x)=8.1883 -8.1918≈-0.0035. Wait, so slightly negative. Hmm, but according to the exact equation, x= arctan(4/3)/8 should satisfy the equation. Wait, perhaps due to calculator precision?Wait, the equation we had was:sin(6x + arctan(9/2)) = sin(2x + arctan(7/6))From the second case, we had:6x + arctan(9/2) = π - (2x + arctan(7/6)) + 2πnWhich led to x= [π - arctan(7/6) - arctan(9/2) +2πn]/8But we found that π - arctan(7/6) - arctan(9/2)=arctan(4/3). Therefore,x= [arctan(4/3) +2πn]/8.Thus, for n=0, x= arctan(4/3)/8≈0.1159 radians. So plugging this into the original equation, it should satisfy.But when I calculated f(x)= left - right≈-0.0035. Close to zero, but not exactly zero. So maybe due to calculator error or perhaps the exact solution requires more precise computation.Alternatively, perhaps this is due to the phase shift approximations. Alternatively, maybe I made an error in deriving the expressions. Let me verify the steps again.Original equation: 2 sin(6x) +9 cos(6x)=6 sin(2x)+7 cos(2x)Expressed as √85 sin(6x + arctan(9/2))=√85 sin(2x + arctan(7/6))Divided by √85: sin(6x + arctan(9/2))=sin(2x + arctan(7/6))Therefore, either:1. 6x + arctan(9/2)=2x + arctan(7/6)+2πnOr2. 6x + arctan(9/2)=π - (2x + arctan(7/6)) +2πnWhich gives solutions:1. 4x= arctan(7/6)-arctan(9/2)+2πn2. 8x=π - arctan(7/6)-arctan(9/2)+2πnThen, as we found, π - arctan(7/6)-arctan(9/2)=arctan(4/3). Therefore, case 2 becomes 8x= arctan(4/3)+2πn, so x=(arctan(4/3)+2πn)/8.Therefore, for n=0, x=arctan(4/3)/8≈0.1159. Which should be the smallest positive solution.But when we plug in x≈0.1159 into the original equation, the left side minus right side is approximately -0.0035, which is very close to zero, but not exactly. So perhaps this is due to calculator precision. Alternatively, perhaps there's an exact solution here. Let's see.Alternatively, maybe using another identity.Alternatively, let me try to subtract the right side from the left and set it to zero:2 sin(6x) +9 cos(6x) -6 sin(2x) -7 cos(2x)=0Alternatively, perhaps express sin(6x) and cos(6x) in terms of sin(2x) and cos(2x). Since 6x is 3*2x. So using multiple-angle formulas.But that might complicate things. Let me recall that sin(3θ)=3 sinθ -4 sin³θ, cos(3θ)=4 cos³θ -3 cosθ. So perhaps:sin(6x)=2 sin(3*2x)=2[3 sin(2x) -4 sin³(2x)]=6 sin(2x)-8 sin³(2x)cos(6x)=cos(3*2x)=4 cos³(2x)-3 cos(2x)So substituting into the equation:2[6 sin(2x)-8 sin³(2x)] +9[4 cos³(2x)-3 cos(2x)] -6 sin(2x)-7 cos(2x)=0Let me compute each term:2*6 sin2x=12 sin2x2*(-8 sin³2x)= -16 sin³2x9*4 cos³2x=36 cos³2x9*(-3 cos2x)= -27 cos2xThen subtract 6 sin2x and subtract7 cos2x:So total:12 sin2x -16 sin³2x +36 cos³2x -27 cos2x -6 sin2x -7 cos2x=0Simplify:(12 sin2x -6 sin2x) + (-16 sin³2x) + (36 cos³2x) + (-27 cos2x -7 cos2x)=0Which is:6 sin2x -16 sin³2x +36 cos³2x -34 cos2x=0Hmm, this seems more complicated. Maybe factor terms:Let me factor terms by sin2x and cos2x:6 sin2x -16 sin³2x +36 cos³2x -34 cos2x=0Group sin terms and cos terms:sin2x(6 -16 sin²2x) + cos2x(36 cos²2x -34)=0Alternatively, factor 2:2[3 sin2x -8 sin³2x +18 cos³2x -17 cos2x]=0Still complicated. Maybe use the identity cos²2x=1 -sin²2x:So 36 cos³2x=36 cos2x (1 - sin²2x)Thus:sin2x(6 -16 sin²2x) +36 cos2x (1 - sin²2x) -34 cos2x=0= sin2x(6 -16 sin²2x) +36 cos2x -36 cos2x sin²2x -34 cos2x=0= sin2x(6 -16 sin²2x) +2 cos2x -36 cos2x sin²2x=0Still messy. Maybe not helpful.Alternatively, let me consider writing the original equation in terms of sin and cos of 2x, then use substitution. Let me set y=2x. Then 6x=3y. So equation becomes:2 sin(3y) +9 cos(3y)=6 sin y +7 cos ySo substitute y=2x, and solve for y.So equation: 2 sin3y +9 cos3y -6 siny -7 cosy=0Now, express sin3y and cos3y in terms of siny and cosy.Using identities:sin3y=3 siny -4 sin³ycos3y=4 cos³y -3 cosySo substituting:2(3 siny -4 sin³y) +9(4 cos³y -3 cosy) -6 siny -7 cosy=0Expand:6 siny -8 sin³y +36 cos³y -27 cosy -6 siny -7 cosy=0Simplify:(6 siny -6 siny) + (-8 sin³y) +36 cos³y + (-27 cosy -7 cosy)=0So:-8 sin³y +36 cos³y -34 cosy=0Factor:-8 sin³y +36 cos³y -34 cosy=0Hmm, this is similar to previous equation.Alternatively, divide both sides by cosy (assuming cosy ≠0):-8 sin³y / cosy +36 cos²y -34=0But -8 sin³y / cosy = -8 siny * tan²y. Not helpful.Alternatively, write everything in terms of siny:Since cos²y=1 - sin²y,36 cos³y=36 cosy (1 - sin²y)So:-8 sin³y +36 cosy (1 - sin²y) -34 cosy=0= -8 sin³y +36 cosy -36 cosy sin²y -34 cosy=0= -8 sin³y +2 cosy -36 cosy sin²y=0Factor terms:= -8 sin³y -36 cosy sin²y +2 cosy=0Factor sin²y:= sin²y (-8 siny -36 cosy) +2 cosy=0Hmm, still not helpful. Maybe this approach isn't working. Let's try another substitution.Let me consider writing the equation as:2 sin3y +9 cos3y =6 siny +7 cosyLet me think of this as a linear combination of sin3y, cos3y, siny, cosy. Perhaps using harmonic addition.Alternatively, express both sides in terms of sin and cos of same angle.Alternatively, use the method of auxiliary angles.But maybe we can write left side as A sin(3y + α) and right side as B sin(y + β). But I don't think that helps directly.Alternatively, move all terms to left side:2 sin3y +9 cos3y -6 siny -7 cosy=0Let me see if I can factor this expression.Hmm, not obvious. Maybe consider writing sin3y in terms of siny and see if terms can be grouped.Recall sin3y=3 siny -4 sin³y. So substitute:2(3 siny -4 sin³y) +9 cos3y -6 siny -7 cosy=0=6 siny -8 sin³y +9 cos3y -6 siny -7 cosy=0= -8 sin³y +9 cos3y -7 cosy=0But this again leads to similar equation as before.Alternatively, express cos3y in terms of cosy:cos3y=4 cos³y -3 cosy. So:-8 sin³y +9(4 cos³y -3 cosy) -7 cosy=0= -8 sin³y +36 cos³y -27 cosy -7 cosy=0= -8 sin³y +36 cos³y -34 cosy=0Same as before.Alternatively, express sin³y in terms of (1 - cos²y)siny:sin³y = siny (1 - cos²y)So:-8 siny (1 - cos²y) +36 cos³y -34 cosy=0= -8 siny +8 siny cos²y +36 cos³y -34 cosy=0Still complicated.Alternatively, maybe substitute t=cosy, then siny=√(1 -t²). But this will lead to a complicated equation. Let me try.Set t=cosy. Then siny=√(1 -t²). The equation:-8 [ (1 -t²)^(3/2) ] +36 t³ -34 t=0This seems even worse. Maybe not helpful.Alternatively, perhaps assume that y is small. For the smallest positive root, y=2x, and x≈0.1159, so y≈0.2318 radians, which is about 13.3 degrees. So small angle. Maybe use Taylor series expansion around y=0.Compute left side:2 sin3y +9 cos3y -6 siny -7 cosyExpand each term in Taylor series up to, say, y^3.sin3y ≈3y - ( (3y)^3 )/6=3y -27y³/6=3y -4.5y³cos3y≈1 - ( (3y)^2 )/2 + ( (3y)^4 )/24≈1 -4.5y² +10.125y^4 (but since we go up to y³, maybe truncate earlier)Similarly, siny≈y - y³/6, cosy≈1 -y²/2.Thus:Left side:2*(3y -4.5y³) +9*(1 -4.5y²) -6*(y - y³/6) -7*(1 -y²/2)Compute term by term:2*(3y -4.5y³)=6y -9y³9*(1 -4.5y²)=9 -40.5y²-6*(y - y³/6)= -6y + y³-7*(1 -y²/2)= -7 +3.5y²Combine all terms:6y -9y³ +9 -40.5y² -6y +y³ -7 +3.5y²Simplify:(6y -6y) + (-9y³ +y³) + (9 -7) + (-40.5y² +3.5y²)=0 -8y³ +2 -37y²So left side≈ -8y³ -37y² +2Set equal to zero:-8y³ -37y² +2=0Solve for y. For small y, the dominant term is 2, then the cubic and quadratic terms. Let's write as:8y³ +37y²=2Approximate solution. Let's try y≈0.23:8*(0.23)^3 +37*(0.23)^2≈8*0.012167 +37*0.0529≈0.0973 +1.9573≈2.0546>2. Close.At y=0.23: left≈2.0546>2At y=0.22:8*(0.22)^3 +37*(0.22)^2≈8*0.010648 +37*0.0484≈0.0852 +1.7908≈1.876<2So between y=0.22 and y=0.23.We can use linear approximation.At y=0.22: f(y)=1.876At y=0.23: f(y)=2.0546We need to find y such that f(y)=2. The difference between y=0.22 and y=0.23:Δy=0.01, Δf=2.0546 -1.876=0.1786Need Δf=2 -1.876=0.124. So fraction=0.124 /0.1786≈0.694. Therefore, y≈0.22 +0.01*0.694≈0.22694Thus, approximate y≈0.22694 radians. Then x=y/2≈0.11347 radians.But this contradicts the previous exact solution of x≈0.1159. Hmm.Alternatively, maybe the Taylor expansion is not accurate enough. Since we truncated higher order terms, the approximation might not be precise.Alternatively, use Newton-Raphson method on the original equation.Define f(y)=2 sin3y +9 cos3y -6 siny -7 cosyWe need to solve f(y)=0.Take y≈0.23 radians.Compute f(0.23):sin3*0.23=sin0.69≈0.6367, cos3*0.69≈cos0.69≈0.7684Left side:2*0.6367 +9*0.7684≈1.2734 +6.9156≈8.1890Right side:6*sin0.23 +7*cos0.23≈6*0.2280 +7*0.9736≈1.368 +6.8152≈8.1832So f(y)=8.1890 -8.1832≈0.0058.Compute f(0.23)=0.0058.Compute derivative f’(y)=6 cos3y -27 sin3y -6 cosy +7 sinyAt y=0.23:cos3*0.23=cos0.69≈0.7684sin3*0.69≈0.6367cos0.23≈0.9736sin0.23≈0.2280So f’(0.23)=6*0.7684 -27*0.6367 -6*0.9736 +7*0.2280Compute each term:6*0.7684≈4.6104-27*0.6367≈-17.1909-6*0.9736≈-5.84167*0.2280≈1.5960Total≈4.6104 -17.1909 -5.8416 +1.5960≈(4.6104 +1.5960) - (17.1909 +5.8416)≈6.2064 -23.0325≈-16.8261Thus, Newton-Raphson update: y1 = y0 - f(y0)/f’(y0)=0.23 -0.0058 / (-16.8261)=0.23 +0.000345≈0.230345 radians.Compute f(0.230345):3y=0.691035 radians.sin(0.691035)≈0.6373, cos≈0.7706Left side:2*0.6373 +9*0.7706≈1.2746 +6.9354≈8.2100Right side:6*sin(0.230345) +7*cos(0.230345)sin(0.230345)≈0.2288, cos≈0.97356*0.2288≈1.3728, 7*0.9735≈6.8145Total≈1.3728 +6.8145≈8.1873Thus, f(y)=8.2100 -8.1873≈0.0227. Wait, this is worse. Hmm, maybe due to higher derivative?Alternatively, perhaps I made an error in calculations. Let me verify:Wait, previous f(y)=0.0058 at y=0.23. Then derivative f’≈-16.8261. So next estimate is y1=0.23 +0.0058/16.8261≈0.23 +0.000345≈0.230345. Then compute f(y1)=?Wait, 3y1=0.691035 radians. So sin(0.691035). Let's use calculator:sin(0.691035)= approx 0.6367 (since sin(0.69)= approx 0.6367). Similarly, cos(0.691035)= approx 0.7711.Left side:2*0.6367=1.2734, 9*0.7711=6.9399. Total≈8.2133.Right side:y1=0.230345. sin(y1)=sin(0.230345)≈0.2290, cos(y1)=cos(0.230345)≈0.9734.6*0.2290≈1.374, 7*0.9734≈6.8138. Total≈8.1878.Thus, f(y1)=8.2133 -8.1878≈0.0255. Wait, that's even higher. That's strange. Newton-Raphson should converge, but maybe due to the derivative being large negative, the function is very steep.Wait, perhaps my calculator approximations are not precise enough. Alternatively, perhaps the function has multiple roots nearby, making Newton-Raphson unstable here. Alternatively, use a different method.Alternatively, use the secant method. Between y=0.22 and y=0.23:f(0.22)= previous approx when using Taylor series gives f(y)=2.0 -8y³ -37y². Wait, no, earlier Taylor approximation was for f(y)= left - right. Let me instead compute actual f(0.22):y=0.22, 3y=0.66 radians.sin(0.66)≈0.6131, cos≈0.7900.Left side:2*0.6131 +9*0.7900≈1.2262 +7.110≈8.3362Right side:6*sin(0.22)+7*cos(0.22)≈6*0.2182 +7*0.9759≈1.3092 +6.8313≈8.1405f(0.22)=8.3362 -8.1405≈0.1957.Previously, at y=0.23, f(y)=0.0058.Wait, so between y=0.22 (f=0.1957) and y=0.23 (f=0.0058). Let's apply the secant method.Slope=(0.0058 -0.1957)/(0.23 -0.22)= (-0.1899)/0.01=-18.99Next estimate: y=0.23 -0.0058*(-18.99)≈0.23 +0.110≈0.34. Wait, that can't be right. Wait, no, secant method formula is:y_{n+1}=y_n - f(y_n)*(y_n -y_{n-1})/(f(y_n) -f(y_{n-1}))So with y0=0.23, f(y0)=0.0058; y1=0.22, f(y1)=0.1957.Then y2=0.23 -0.0058*(0.23 -0.22)/(0.0058 -0.1957)=0.23 -0.0058*(0.01)/(-0.1899)=0.23 +0.0058*0.01/0.1899≈0.23 +0.000306≈0.230306Compute f(y2)=f(0.230306). As before, approx 0.0255. Hmm, this is not working. Perhaps the function is very flat here, making secant method unreliable. Alternatively, use linear interpolation.We have two points: y=0.22, f=0.1957; y=0.23, f=0.0058. Assume linearity between these points.The desired f=0. The x-intercept is given by:y=0.22 + (0 -0.1957)*(0.23 -0.22)/(0.0058 -0.1957)=0.22 + (-0.1957)*(0.01)/(-0.1899)=0.22 +0.1957*0.01/0.1899≈0.22 +0.0103≈0.2303So y≈0.2303. Then x=y/2≈0.11515 radians. Which is close to our earlier exact solution of≈0.1159. The difference is due to linear approximation.Thus, the actual root is approximately 0.1152 to 0.1159 radians. Given the exact expression x=arctan(4/3)/8≈0.1159, which is very close to these approximations. Therefore, the smallest positive root is x=arctan(4/3)/8.But let's confirm if this is indeed the correct exact solution.We had earlier that 8x=arctan(4/3). Therefore, x=arctan(4/3)/8. Let's verify this satisfies the original equation.Compute left side:2 sin(6x)+9 cos(6x). With x=arctan(4/3)/8, 6x= (6/8) arctan(4/3)= (3/4) arctan(4/3).Similarly, let's compute sin(6x + arctan(9/2)) and see if it equals sin(2x + arctan(7/6)).Wait, from our earlier steps, we derived that x=arctan(4/3)/8 is a solution. Since the derivation was algebraically correct, except for possible periodicity issues, but since we're looking for the smallest positive solution, n=0 gives the first solution.Therefore, the exact smallest positive solution is x=arctan(4/3)/8. Converting this to a decimal, arctan(4/3)≈0.927295218, so x≈0.927295218/8≈0.1159119 radians. To check if this is indeed the smallest positive solution, we can see from the earlier analysis that the next possible solution from the first case is x≈1.448 radians, which is larger. Thus, x≈0.1159 is indeed the smallest positive root.Therefore, the answer is x= arctan(4/3)/8. However, the problem asks for the smallest positive root, and it may expect an exact form or a numerical value. Since arctan(4/3) doesn't correspond to a standard angle, likely the answer should be expressed as arctan(4/3)/8, but maybe the problem expects a numerical value. Let's check if arctan(4/3)/8 is the exact answer.Alternatively, perhaps there's a way to express this more elegantly. But I think it's acceptable to leave it as arctan(4/3)/8. Alternatively, since arctan(4/3)=π/2 - arctan(3/4), but that might not help.Alternatively, in terms of logarithms, but that's more complicated. So likely, the exact answer is x= (1/8) arctan(4/3). Therefore, boxed answer is boxed{dfrac{1}{8} arctanleft(dfrac{4}{3}right)}.Alternatively, if a numerical value is required, approximately 0.1159 radians, which is roughly 6.64 degrees. But since the problem doesn't specify, and mathematical problems often prefer exact forms, I think the exact expression is appropriate.Final AnswerThe smallest positive root is boxed{dfrac{1}{8} arctanleft(dfrac{4}{3}right)}.</think>To find the smallest positive root of the equation:[ 2 sin(6x) + 9 cos(6x) = 6 sin(2x) + 7 cos(2x) ]we start by rewriting both sides using the amplitude-phase form. The left side (2 sin(6x) + 9 cos(6x)) can be written as (sqrt{85} sin(6x + arctan(9/2))), and the right side (6 sin(2x) + 7 cos(2x)) can be written as (sqrt{85} sin(2x + arctan(7/6))). Dividing both sides by (sqrt{85}), we get:[ sin(6x + arctan(9/2)) = sin(2x + arctan(7/6)) ]The solutions to (sin alpha = sin beta) are (alpha = beta + 2pi n) or (alpha = pi - beta + 2pi n). This gives us two cases:1. (6x + arctan(9/2) = 2x + arctan(7/6) + 2pi n)2. (6x + arctan(9/2) = pi - (2x + arctan(7/6)) + 2pi n)For the first case, simplifying gives:[ 4x = arctan(7/6) - arctan(9/2) + 2pi n ]For the second case, simplifying gives:[ 8x = pi - arctan(7/6) - arctan(9/2) + 2pi n ]We need to find the smallest positive (x). Using the identity for the sum of arctangents, we find:[ arctan(7/6) + arctan(9/2) = pi - arctan(4/3) ]Thus, the second case simplifies to:[ 8x = arctan(4/3) + 2pi n ]For (n = 0), the smallest positive solution is:[ x = frac{arctan(4/3)}{8} ]Therefore, the smallest positive root of the equation is:[boxed{dfrac{1}{8} arctanleft(dfrac{4}{3}right)}],<think>Okay, let me try to figure out how to solve this equation: 2 sin(6x) + 9 cos(6x) = 6 sin(2x) + 7 cos(2x). Hmm, so we have trigonometric functions with different arguments, 6x and 2x. I need to find the smallest positive root. Alright, step by step.First, maybe I can rewrite the equation in a way that relates the arguments. Since 6x is three times 2x, maybe there's a trigonometric identity that connects sin(3θ) and cos(3θ) to sin(θ) and cos(θ). Let me recall those identities. The triple angle formulas:sin(3θ) = 3 sinθ - 4 sin³θcos(3θ) = 4 cos³θ - 3 cosθBut here we have sin(6x) and cos(6x), which can be written as sin(3*2x) and cos(3*2x). So maybe if I let θ = 2x, then the equation becomes:2 sin(3θ) + 9 cos(3θ) = 6 sinθ + 7 cosθThat seems manageable. Let me substitute θ = 2x. So our equation is:2 sin(3θ) + 9 cos(3θ) = 6 sinθ + 7 cosθNow, let's replace sin(3θ) and cos(3θ) using the triple angle formulas.So:2*(3 sinθ - 4 sin³θ) + 9*(4 cos³θ - 3 cosθ) = 6 sinθ + 7 cosθLet me expand that:Left side: 2*3 sinθ - 2*4 sin³θ + 9*4 cos³θ - 9*3 cosθWhich simplifies to:6 sinθ - 8 sin³θ + 36 cos³θ - 27 cosθRight side: 6 sinθ + 7 cosθSo, moving everything to the left side:6 sinθ - 8 sin³θ + 36 cos³θ - 27 cosθ - 6 sinθ - 7 cosθ = 0Simplify term by term:6 sinθ - 6 sinθ = 0-8 sin³θ remains36 cos³θ remains-27 cosθ -7 cosθ = -34 cosθSo overall:-8 sin³θ + 36 cos³θ -34 cosθ = 0Hmm, that's a cubic in sinθ and cosθ. Maybe factor out common terms? Let's see:Looking at coefficients: -8, 36, -34. Hmm, not obvious. Let's see:Maybe factor out a common factor from the first two terms? -8 sin³θ + 36 cos³θ. Hmm, -8 and 36 have a common factor of 4? Let's try:-8 sin³θ + 36 cos³θ = 4*(-2 sin³θ + 9 cos³θ)So then the equation becomes:4*(-2 sin³θ + 9 cos³θ) -34 cosθ = 0But not sure if that helps. Alternatively, perhaps factor cosθ from the last two terms:36 cos³θ -34 cosθ = cosθ(36 cos²θ -34)So then the entire equation is:-8 sin³θ + cosθ(36 cos²θ -34) = 0Still complicated. Maybe express everything in terms of sinθ or cosθ. Let's try expressing cos²θ as 1 - sin²θ.So cosθ(36(1 - sin²θ) -34) = cosθ(36 -36 sin²θ -34) = cosθ(2 -36 sin²θ)Thus, equation becomes:-8 sin³θ + cosθ*(2 -36 sin²θ) = 0Hmm. That's:-8 sin³θ -36 sin²θ cosθ + 2 cosθ = 0Not sure. Maybe factor terms with sinθ and cosθ. Let's group terms:-8 sin³θ -36 sin²θ cosθ + 2 cosθ = 0Factor sin²θ from the first two terms:sin²θ(-8 sinθ -36 cosθ) + 2 cosθ = 0Hmm, maybe:sin²θ*(-8 sinθ -36 cosθ) + 2 cosθ = 0Not sure. Let's see if there's another approach.Alternatively, perhaps use substitution variables. Let’s set t = θ, so we have the equation in terms of t. Maybe divide both sides by cos^3θ? Hmm, but that might complicate. Alternatively, use substitution like u = tanθ, since that can convert sin and cos into tan.Let me try that. Let u = tanθ, so sinθ = u / sqrt(1 + u²), cosθ = 1 / sqrt(1 + u²). But this might get messy with the cubic terms. Let me see:But let me check the equation again:-8 sin³θ + 36 cos³θ -34 cosθ = 0Let’s divide both sides by cos³θ (assuming cosθ ≠ 0, which we can check later):-8 (sinθ/cosθ)^3 + 36 -34 (1/cos²θ) = 0But 1/cos²θ = 1 + tan²θ, so:-8 tan³θ + 36 -34(1 + tan²θ) = 0Let’s let u = tanθ:Then:-8 u³ + 36 -34 -34 u² = 0Simplify:-8 u³ -34 u² + 2 = 0Multiply both sides by -1:8 u³ +34 u² -2 = 0So, 8u³ +34u² -2 = 0Hmm, solving this cubic equation for u. Let me see if I can factor this.Let’s try rational roots. The possible rational roots are ±1, ±2, ±1/2, ±1/4, etc., factors of 2 over factors of 8.Testing u=1: 8 +34 -2 =40 ≠0u= -1: -8 +34 -2=24≠0u=1/2: 8*(1/8) +34*(1/4) -2=1 +8.5 -2=7.5≠0u= -1/2: 8*(-1/2)^3 +34*(-1/2)^2 -2=8*(-1/8)+34*(1/4)-2= -1 +8.5 -2=5.5≠0u=2: 8*8 +34*4 -2=64 +136 -2=198≠0u= -2: 8*(-8) +34*4 -2= -64 +136 -2=70≠0u=1/4: 8*(1/64) +34*(1/16) -2=0.125 +2.125 -2=0.25≠0u=-1/4: 8*(-1/64) +34*(1/16) -2= -0.125 +2.125 -2=0Wait! u=-1/4 gives:8*(-1/4)^3 +34*(-1/4)^2 -2=8*(-1/64)+34*(1/16)-2= -0.125 +2.125 -2=0Yes! So u=-1/4 is a root. Therefore, (u +1/4) is a factor.Let’s perform polynomial division or use synthetic division.Given the cubic equation 8u³ +34u² -2 =0, and we found a root at u=-1/4.So let's factor out (u +1/4). Let me use polynomial division.Divide 8u³ +34u² -2 by (u +1/4). Let's write the coefficients:8 | 34 | 0 | -2Wait, but the cubic is 8u³ +34u² +0u -2.Using synthetic division with root -1/4:Set up coefficients: 8, 34, 0, -2Multiply and add:Bring down 8.Multiply 8 by -1/4: -2Add to next coefficient:34 + (-2)=32Multiply 32 by -1/4: -8Add to next coefficient:0 + (-8)= -8Multiply -8 by -1/4: 2Add to last coefficient: -2 +2=0So the quotient polynomial is 8u² +32u -8Therefore, the cubic factors as (u +1/4)(8u² +32u -8)=0So the roots are u=-1/4 and roots of 8u² +32u -8=0.Let me solve the quadratic:8u² +32u -8 =0Divide by 8: u² +4u -1=0Solutions: u=(-4 ±√(16 +4))/2=(-4 ±√20)/2=(-4 ±2√5)/2= -2 ±√5So the roots are u=-1/4, u=-2+√5≈-2+2.236≈0.236, and u=-2-√5≈-4.236.So tanθ = u = -1/4, tanθ= -2+√5, tanθ=-2-√5.So θ = arctan(-1/4) +kπ, θ= arctan(-2+√5)+kπ, θ=arctan(-2-√5)+kπ.But since θ=2x, and we need to find the smallest positive x, we need to find the smallest positive θ satisfying the equation and then divide by 2.Let me analyze each solution.First, tanθ = -1/4. The principal solution is θ = arctan(-1/4), which is in the fourth quadrant. But since tangent has a period of π, the general solution is θ = -arctan(1/4) +kπ. To get positive θ, we can take k=1: θ= π - arctan(1/4). Also, k=2: θ=2π - arctan(1/4), etc. So the smallest positive θ from this root is π - arctan(1/4).Second, tanθ= -2+√5≈0.236. So arctan(-2+√5)= arctan(approx 0.236). The principal value is in the first quadrant because the argument is positive. Wait, hold on: -2+√5 is approximately -2 +2.236=0.236, which is positive. So tanθ=0.236, so θ= arctan(0.236) +kπ. The principal value is in the first quadrant. So the smallest positive θ here is arctan(-2+√5)= arctan(√5 -2). Since √5≈2.236, √5 -2≈0.236.Third, tanθ= -2 -√5≈-4.236. The principal solution is θ= arctan(-4.236), which is in the fourth quadrant. The positive solutions would be θ= π + arctan(-4.236)= π - arctan(4.236). So θ≈π -1.34≈1.80 radians. But let's check exact values.So now, to find the smallest positive θ, we need to compare the three possible solutions:1. θ1= π - arctan(1/4)2. θ2= arctan(√5 -2)3. θ3= π - arctan(2 +√5)Wait, because tanθ= -2 -√5 is negative, so θ= -arctan(2 +√5) +kπ. So positive solutions would be θ= π - arctan(2 +√5) +kπ. But arctan(2 +√5) is a specific value. Let me compute arctan(√5 -2) and arctan(2 +√5).Note that √5 ≈2.236, so √5 -2≈0.236 and 2 +√5≈4.236.But interestingly, arctan(√5 -2) and arctan(2 +√5) might have a relationship. Let me check:Let’s compute tan(π/8):tan(π/8)=√2 -1≈0.414. Hmm, √5 -2≈0.236, which is smaller. Maybe arctan(√5 -2) is π/10 or something? Wait, not sure. Alternatively, perhaps there's a complementary angle.Wait, let's recall that tan(θ) = √5 -2. Let's square it:(√5 -2)^2=5 -4√5 +4=9 -4√5≈9 -8.944≈0.056. Not helpful.Alternatively, perhaps there is an identity. Let me consider θ= arctan(√5 -2). Let’s compute tan(2θ):tan(2θ)=2 tanθ/(1 - tan²θ)=2*(√5 -2)/(1 - (√5 -2)^2)Compute denominator:1 - (5 -4√5 +4)=1 -9 +4√5= -8 +4√5So tan(2θ)=2*(√5 -2)/(-8 +4√5)= multiply numerator and denominator by (-1):2*(2 -√5)/(8 -4√5)= factor denominator: 4*(2 -√5)So numerator: 2*(2 -√5); denominator:4*(2 -√5)So tan(2θ)= (2*(2 -√5))/(4*(2 -√5))=1/2. Therefore, tan(2θ)=1/2. Therefore, 2θ= arctan(1/2). So θ= (1/2) arctan(1/2). Hmm, not a standard angle, but useful to know.Similarly, for tanθ=2 +√5, let's compute tan(2θ):tan(2θ)=2*(2 +√5)/(1 - (2 +√5)^2)Denominator:1 - (4 +4√5 +5)=1 -9 -4√5= -8 -4√5So tan(2θ)=2*(2 +√5)/(-8 -4√5)= multiply numerator and denominator by (-1):2*(-2 -√5)/(8 +4√5)= factor denominator:4*(2 +√5)Numerator: -2*(2 +√5)So tan(2θ)= (-2*(2 +√5))/(4*(2 +√5))= -2/4= -1/2. Therefore, tan(2θ)= -1/2. Therefore, 2θ= arctan(-1/2) +kπ. So θ= (1/2)(- arctan(1/2) +kπ). For positive θ, take k=1: θ=(π - arctan(1/2))/2.But maybe this is overcomplicating. Let's get back to the solutions.We have three possible θ:1. θ1= π - arctan(1/4)2. θ2= arctan(√5 -2)≈ arctan(0.236)≈0.232 radians3. θ3= π - arctan(2 +√5)≈π - arctan(4.236)≈π -1.34≈1.80 radiansSo among these, θ2≈0.232 is the smallest positive solution. Then θ1≈π -0.245≈2.896, and θ3≈1.80. So the smallest positive θ is θ2≈0.232 radians.But we need to check if these solutions actually satisfy the original equation, because sometimes when we do substitutions and square things or divide by variables, we might introduce extraneous solutions. But since we didn't square, but we did divide by cos^3θ, assuming cosθ≠0. So if cosθ=0, we need to check separately. But when cosθ=0, θ=π/2 +kπ. Let's check if these satisfy the original equation.Original equation: -8 sin³θ +36 cos³θ -34 cosθ=0If cosθ=0, then sinθ=±1. Substitute cosθ=0, sinθ=±1:-8*(±1)^3 +0 -0= -8*(±1)=∓8≠0. So no solution when cosθ=0. So our solutions are valid.Therefore, the smallest positive θ is θ2= arctan(√5 -2)≈0.232 radians. Then, since θ=2x, x=θ/2≈0.232/2≈0.116 radians. But we need to verify if this is indeed the smallest positive solution.But wait, θ3=1.80 radians is larger than θ1≈2.896? Wait, no:Wait θ3≈1.80, θ1≈π -0.245≈2.896. So θ3 is smaller than θ1. So θ2≈0.232, θ3≈1.80, θ1≈2.896. So the order is θ2 < θ3 < θ1. Therefore, the smallest positive θ is θ2≈0.232.But let's confirm numerically. Let me compute θ2= arctan(√5 -2). Let's compute tan(θ2)=√5 -2≈0.236. Let's compute θ2≈ arctan(0.236). Using a calculator, arctan(0.236)≈0.232 radians. So θ2≈0.232, which is the smallest.But let's check θ3= π - arctan(2 +√5). Since tan(arctan(2 +√5))=2 +√5≈4.236. So arctan(4.236)≈1.34 radians. Thus, θ3≈π -1.34≈1.80 radians.So θ2≈0.232, θ3≈1.80, θ1≈2.896.Therefore, the smallest positive θ is θ2≈0.232, so x=θ2/2≈0.116 radians. But let's check if there are smaller positive solutions from the other roots. For example, when tanθ= -1/4, the smallest positive θ is π - arctan(1/4). Let's compute arctan(1/4)≈0.245 radians, so θ1≈π -0.245≈2.896 radians. So x1≈2.896/2≈1.448 radians. Which is larger than x2≈0.116. Similarly, x3≈1.80/2≈0.9 radians. So x3≈0.9, which is larger than x2≈0.116. So indeed, x2≈0.116 is the smallest.But to ensure this is correct, let's plug x≈0.116 into the original equation.Original equation: 2 sin(6x) +9 cos(6x)=6 sin(2x)+7 cos(2x)Compute left side: 2 sin(6*0.116) +9 cos(6*0.116)6x≈0.696 radianssin(0.696)≈0.642, cos(0.696)≈0.766So left≈2*0.642 +9*0.766≈1.284 +6.894≈8.178Right side:6 sin(2*0.116)+7 cos(2*0.116)2x≈0.232 radianssin(0.232)≈0.230, cos(0.232)≈0.973Right≈6*0.230 +7*0.973≈1.38 +6.811≈8.191Close enough, considering the approximations. The slight discrepancy is due to rounding errors. So it seems correct.Alternatively, let's compute exactly using θ= arctan(√5 -2). Let me see:tanθ=√5 -2. Then sinθ= (√5 -2)/sqrt(1 + (√5 -2)^2 )Compute denominator:(√5 -2)^2=5 -4√5 +4=9 -4√5So sqrt(1 +9 -4√5)=sqrt(10 -4√5). Hmm, can this be simplified?Let’s compute 10 -4√5≈10 -8.944≈1.056, so sqrt(1.056)≈1.027. Not helpful. But maybe express sinθ and cosθ in exact terms.But perhaps for the purposes of finding the root, we can just express x as (1/2) arctan(√5 -2). But the problem asks for the smallest positive root. It might expect an exact expression, perhaps in terms of π or something. Wait, but is arctan(√5 -2) a standard angle?Earlier, we saw that tan(2θ)=1/2 when tanθ=√5 -2. So 2θ= arctan(1/2). Therefore, θ= (1/2) arctan(1/2). Therefore, the solution x=θ/2= (1/4) arctan(1/2). But arctan(1/2) is not a standard angle. Alternatively, maybe there's another identity.Wait, but let's recall that arctan(1/2) can be expressed in terms of other arctangent expressions, but I don't think it's a standard angle. Therefore, perhaps the answer is expressed in terms of arctan(√5 -2) or (1/2) arctan(1/2). Alternatively, maybe it relates to π/10 or some other fraction. Let me check.Alternatively, let's recall that tan(π/10)≈0.3249, which is larger than tanθ=0.236. Similarly, tan(π/12)≈0.2679, still larger. tan(π/16)≈0.1989, which is less than 0.236. So arctan(0.236) is between π/16≈0.196 and π/12≈0.261. So approximately 0.232 radians. Not a standard angle. Therefore, the exact answer is x= (1/2) arctan(√5 -2).Alternatively, using the earlier substitution, since tan(2θ)=1/2, so 2θ= arctan(1/2), so θ= (1/2) arctan(1/2), and x=θ/2= (1/4) arctan(1/2). But both forms are acceptable. However, the problem says "Find the smallest positive root", and it might expect an exact value. Let's see if arctan(√5 -2) can be expressed in terms of π fractions.Alternatively, perhaps we can use another method to solve the original equation, leading to a nicer expression.Let me revisit the original equation:2 sin(6x) +9 cos(6x) =6 sin(2x) +7 cos(2x)Maybe instead of expanding sin(6x) and cos(6x), we can write both sides as sinusoidal functions.The left side: 2 sin(6x) +9 cos(6x). This can be written as A sin(6x + φ), where A=√(2² +9²)=√(4 +81)=√85≈9.2195, and φ= arctan(9/2). Similarly, the right side:6 sin(2x) +7 cos(2x)= B sin(2x + ψ), where B=√(6² +7²)=√85≈9.2195, and ψ= arctan(7/6).Wait, interesting! Both sides have the same amplitude √85. So the equation becomes:√85 sin(6x + φ) = √85 sin(2x + ψ)Divide both sides by √85:sin(6x + φ) = sin(2x + ψ)Which implies that either:1. 6x + φ = 2x + ψ +2πk, or2. 6x + φ = π - (2x + ψ) +2πkfor some integer k.Let’s write these two cases:Case 1:6x + φ =2x + ψ +2πk ⇒4x= ψ -φ +2πk ⇒x=(ψ -φ +2πk)/4Case 2:6x + φ= π -2x -ψ +2πk ⇒8x= π -φ -ψ +2πk ⇒x=(π -φ -ψ +2πk)/8Now, φ= arctan(9/2), ψ= arctan(7/6)We need to compute ψ -φ and π -φ -ψ.First, compute ψ -φ= arctan(7/6) - arctan(9/2)Recall that arctan a - arctan b= arctan((a -b)/(1 +ab)) if certain conditions hold.But here, since 7/6≈1.166 and 9/2=4.5, arctan(7/6) is in the first quadrant, arctan(9/2) is also in the first quadrant, so the difference is arctan(7/6) - arctan(9/2). Let me use the formula:arctan a - arctan b= arctan((a -b)/(1 +ab)) + adjustment if necessary.But since a < b here (7/6≈1.166 <9/2=4.5), the result will be negative. Therefore, arctan(7/6) - arctan(9/2)= - [arctan(9/2) - arctan(7/6)]Alternatively, using the formula:tan(arctan a - arctan b)= (a -b)/(1 +ab)Thus, tan(ψ -φ)= (7/6 -9/2)/(1 + (7/6)(9/2))= ( (7/6 -27/6) )/(1 +63/12)= (-20/6)/(1 +21/4)= (-10/3)/(25/4)= (-10/3)*(4/25)= -40/75= -8/15Therefore, ψ -φ= arctan(-8/15). So the angle whose tangent is -8/15, which is in the fourth quadrant. But since ψ and φ are both positive, ψ -φ is negative. Therefore, ψ -φ= - arctan(8/15). Thus, in case 1:x=(ψ -φ +2πk)/4= (- arctan(8/15) +2πk)/4To find the smallest positive x, take k=1:x= (2π - arctan(8/15))/4≈(6.283 -0.4899)/4≈5.793/4≈1.448 radians, which matches our previous result for x1≈1.448.For case 2:x=(π -φ -ψ +2πk)/8Compute π -φ -ψ.φ= arctan(9/2), ψ=arctan(7/6)So π -φ -ψ= π - arctan(9/2) - arctan(7/6)Again, using tan(arctan a + arctan b)= (a +b)/(1 -ab)But let's compute tan(π -φ -ψ)= tan(π - (φ +ψ))= -tan(φ +ψ)Compute tan(φ +ψ)= (9/2 +7/6)/(1 - (9/2)(7/6))= ( (27/6 +7/6) )/(1 -63/12)= (34/6)/( -51/12)= (17/3)/(-17/4)= -4/3Therefore, tan(φ +ψ)= -4/3. Therefore, tan(π -φ -ψ)= -tan(φ +ψ)=4/3Thus, π -φ -ψ= arctan(4/3). Because tan(arctan(4/3))=4/3, and since π -φ -ψ is in the first quadrant (since φ and ψ are both in the first quadrant, their sum is less than π/2 + π/2=π, so π - their sum is positive). But wait, arctan(4/3) is in the first quadrant. Wait, π -φ -ψ= arctan(4/3). Is this correct? Let's check:If tan(π -φ -ψ)=4/3, then π -φ -ψ= arctan(4/3) +nπ. But since φ and ψ are both positive and less than π/2, their sum is less than π, so π -φ -ψ is between π -0 -0=π and π -π/2 -π/2=0. Wait, actually, if φ and ψ are both less than π/2, their sum is less than π, so π - their sum is between 0 and π. But since tan(π -φ -ψ)=4/3>0, the angle must be in the first or third quadrant. But π -φ -ψ is between 0 and π, so it's in the first or second quadrant. But tan is positive in first and third, so it must be in the first quadrant. Therefore, π -φ -ψ= arctan(4/3). Therefore, π -φ -ψ= arctan(4/3). Therefore, case 2:x=(arctan(4/3) +2πk)/8So the smallest positive x is when k=0:x= arctan(4/3)/8≈ arctan(1.333)/8≈0.927/8≈0.1159 radians≈0.116 radians, which matches our previous result for x2≈0.116. So that's the smallest positive solution.Therefore, the exact value is x= arctan(4/3)/8.But arctan(4/3) can be related to our previous result. Since we had earlier that θ=2x= arctan(√5 -2). So 2x= arctan(√5 -2) ⇒x= (1/2) arctan(√5 -2). But we also have x= arctan(4/3)/8. Are these expressions equivalent?Let me check if arctan(4/3)/8= (1/2) arctan(√5 -2).Let’s compute both sides:First, note that from the previous substitution, tan(2θ)=1/2 where θ= arctan(√5 -2). So 2θ= arctan(1/2). Therefore, θ= (1/2) arctan(1/2). Therefore, x=θ/2= (1/4) arctan(1/2).But we have another expression x= arctan(4/3)/8. Are these the same?Alternatively, let's relate arctan(4/3) and arctan(1/2). Let me see:Let’s suppose that arctan(4/3)=2 arctan(1/2). Let's check:tan(2 arctan(1/2))=2*(1/2)/(1 - (1/2)^2)=1/(1 -1/4)=1/(3/4)=4/3. Yes! Therefore, 2 arctan(1/2)=arctan(4/3). Therefore, arctan(4/3)=2 arctan(1/2). Therefore, x= arctan(4/3)/8= (2 arctan(1/2))/8= arctan(1/2)/4. Which is the same as our previous expression.Similarly, since θ=2x= arctan(√5 -2), we had x= (1/2) arctan(√5 -2). Let's check if arctan(√5 -2)= arctan(1/2)/2.Wait, no. Earlier, we found that tan(2θ)=1/2 where θ= arctan(√5 -2). Therefore, 2θ= arctan(1/2), so θ= (1/2) arctan(1/2). Therefore, x=θ/2= (1/4) arctan(1/2). Which is the same as x= arctan(4/3)/8= (2 arctan(1/2))/8= (1/4) arctan(1/2). So both expressions lead to x= (1/4) arctan(1/2).Alternatively, perhaps express in terms of π? But arctan(1/2) is not a standard angle. Therefore, the exact answer can be written in multiple forms:x= (1/4) arctan(1/2)x= (1/8) arctan(4/3)x= (1/2) arctan(√5 -2)All these are equivalent. However, the problem asks for the smallest positive root. Unless it's a standard angle, which it doesn't seem to be, we need to present the exact expression. However, the problem might expect a numerical value, but the instruction says to "put your final answer within boxed{}", and usually for such problems, an exact form is preferred.But let me check if there is a more simplified exact form. For example, if we use the expression x= (1/4) arctan(1/2), but that might not be simpler. Alternatively, the original equation's symmetry might hint at another approach.Alternatively, recall that when we expressed both sides as √85 sin(6x +φ)=√85 sin(2x +ψ). Therefore, the solutions are given by the cases above. The smallest positive solution comes from case 2 with k=0: x= arctan(4/3)/8. So this is a valid exact form.Alternatively, since arctan(4/3) is an angle whose tan is 4/3, which is a well-known ratio. But I don't think there's a simpler exact form. Therefore, the answer is x= (1/8) arctan(4/3).But let's check if this is equal to (1/2) arctan(√5 -2). Let's compute:Let’s compute arctan(√5 -2). Let’s set θ= arctan(√5 -2). Then tanθ=√5 -2. Let’s compute tan(4θ) to see if it relates to 4/3:First, tan(2θ)=1/2 as established earlier.Then tan(4θ)=2 tan(2θ)/(1 - tan²2θ)=2*(1/2)/(1 -1/4)=1/(3/4)=4/3. Therefore, 4θ= arctan(4/3). Therefore, θ= (1/4) arctan(4/3). But θ= arctan(√5 -2), so arctan(√5 -2)= (1/4) arctan(4/3). Therefore, x=θ/2= (1/8) arctan(4/3). So both expressions are equivalent.Thus, the smallest positive root is x= (1/8) arctan(4/3) or x= (1/4) arctan(1/2), etc. Since the problem says "Find the smallest positive root", and unless a specific form is required, we can choose either. However, arctan(4/3) is perhaps more straightforward as it comes directly from the sinusoidal form.But let me check the initial exact solutions from the substitution method. When we had tanθ=√5 -2, θ=2x, so x= (1/2) arctan(√5 -2). But we also have x= (1/8) arctan(4/3). Are these equal? Let’s see:From tan(4θ)=4/3 where θ= arctan(√5 -2). Wait, 4θ= arctan(4/3), so θ= (1/4) arctan(4/3). But θ= arctan(√5 -2). Therefore, arctan(√5 -2)= (1/4) arctan(4/3). Therefore, x= (1/2) arctan(√5 -2)= (1/2)*(1/4 arctan(4/3))= (1/8) arctan(4/3). So yes, they are equivalent.Therefore, the exact answer is x= (1/8) arctan(4/3). Alternatively, x= (1/4) arctan(1/2). Both are correct, but arctan(4/3) is perhaps more direct.Alternatively, since 4/3 is the tangent of a well-known angle in a 3-4-5 triangle, but it's not a standard angle in terms of π. Therefore, the answer is best left as x= (1/8) arctan(4/3) or another equivalent expression.But let me check if there's a mistake here. Wait, in case 2, we had x=(π -φ -ψ +2πk)/8. Then we found that π -φ -ψ= arctan(4/3). Therefore, the solution is x=(arctan(4/3) +2πk)/8. For k=0, x= arctan(4/3)/8≈0.927/8≈0.116, which is the smallest positive solution. Therefore, the exact value is arctan(4/3)/8. However, since the problem asks for the smallest positive root, this is the exact value. Alternatively, the problem might prefer the answer expressed in terms of π, but unless there's a specific angle that corresponds to arctan(4/3), which there isn't in standard angles, this is the simplest exact form.Alternatively, another approach might yield a different exact form, but I think this is the most straightforward. Therefore, the smallest positive root is x= arctan(4/3)/8.However, the problem might expect the answer in terms of a numerical approximation. Let me compute arctan(4/3)≈0.927295218 radians. Divide by 8:≈0.115911902 radians. To check if this is correct, let's compute 0.115911902*8≈0.927295218, yes. Then convert to degrees if needed, but the question doesn't specify. Since the problem is given in radians, the answer should be in radians.Alternatively, if the problem expects an expression like (1/8) arctan(4/3), which is exact, then that's acceptable. But to check if there's a more simplified exact form, perhaps using multiple-angle identities, but I don't see it. Therefore, I think the answer is x= (1/8) arctan(4/3). But let's check with another method.Wait, going back to the original substitution where we arrived at x= (1/2) arctan(√5 -2). Let's compute arctan(√5 -2). As we saw earlier, this is approximately0.232 radians. Then x=0.232/2≈0.116, which matches. Also, note that arctan(√5 -2)= (1/4) arctan(4/3). Therefore, x= (1/8) arctan(4/3)= (1/2)*(1/4) arctan(4/3)= (1/2)* arctan(√5 -2). Either form is acceptable.But perhaps the answer is better expressed as x= (1/8) arctan(4/3) because it's derived directly from the sinusoidal equation form. However, depending on the problem's expectation, another form might be preferred.Alternatively, the problem might have a clever substitution that leads to a simpler answer, but given the time I've spent and the methods I've tried, I think this is the correct exact form. Therefore, the smallest positive root is x= (1/8) arctan(4/3).But to confirm, let's see if this is equivalent to the earlier solutions. Yes, through multiple substitutions and identities, we saw that all forms are equivalent.Alternatively, perhaps we can rationalize it further. For example, using the identity that arctan(4/3)= arcsin(4/5), since in a 3-4-5 triangle, sin(arctan(4/3))=4/5. Therefore, arctan(4/3)= arcsin(4/5). Therefore, x= (1/8) arcsin(4/5). But this might not be simpler.Alternatively, if we use the expression with arcsin, but again, it's equivalent.In conclusion, the smallest positive root of the equation is x= (1/8) arctan(4/3). Therefore, the answer is boxed{dfrac{1}{8} arctanleft(dfrac{4}{3}right)}.Alternatively, if the problem expects a numerical answer, it would be approximately 0.116 radians. But since the problem is likely expecting an exact answer, the boxed expression is appropriate.Final AnswerThe smallest positive root is boxed{dfrac{1}{8} arctanleft(dfrac{4}{3}right)}.</think>To find the smallest positive root of the equation (2 sin(6x) + 9 cos(6x) = 6 sin(2x) + 7 cos(2x)), we start by using trigonometric identities and substitutions.First, we substitute (theta = 2x), which transforms the equation into:[2 sin(3theta) + 9 cos(3theta) = 6 sin(theta) + 7 cos(theta)]Using the triple angle formulas for sine and cosine, we expand and simplify:[6 sin(theta) - 8 sin^3(theta) + 36 cos^3(theta) - 27 cos(theta) = 6 sin(theta) + 7 cos(theta)]After simplifying, we get:[-8 sin^3(theta) + 36 cos^3(theta) - 34 cos(theta) = 0]We then use the substitution (u = tan(theta)) and solve the resulting cubic equation:[8u^3 + 34u^2 - 2 = 0]Factoring out the root (u = -frac{1}{4}), we solve the quadratic equation and find the roots:[u = -2 + sqrt{5}, quad u = -2 - sqrt{5}]The smallest positive solution for (theta) is (theta = arctan(sqrt{5} - 2)). Converting back to (x), we have:[x = frac{1}{2} arctan(sqrt{5} - 2)]Using trigonometric identities, we find that:[arctan(sqrt{5} - 2) = frac{1}{4} arctanleft(frac{4}{3}right)]Thus, the smallest positive root (x) is:[x = frac{1}{8} arctanleft(frac{4}{3}right)]Therefore, the smallest positive root of the equation is:[boxed{dfrac{1}{8} arctanleft(dfrac{4}{3}right)}]]

🔑:Answer: frac{alpha+beta}{8} approx 0.1159, where alpha=operatorname{arctg} frac{2}{9}, beta=operatorname{arctg} frac{6}{7}.Comment 1. Depending on the choice of the solution method and the inverse trigonometric function, the answer can have a different form. Let's provide several ways to describe alpha and beta:begin{aligned}& alpha=operatorname{arcctg} frac{9}{2}=arcsin frac{2}{sqrt{85}}=arccos frac{9}{sqrt{85}}=frac{pi}{2}-arcsin frac{9}{sqrt{85}}=ldots & beta=operatorname{arcctg} frac{7}{6}=arcsin frac{6}{sqrt{85}}=arccos frac{7}{sqrt{85}}=frac{pi}{2}-arcsin frac{7}{sqrt{85}}=ldotsend{aligned}For example, the answer can be rewritten asfrac{1}{8}left(pi-arcsin frac{9}{sqrt{85}}-arcsin frac{7}{sqrt{85}}right)For convenience of verification, an approximate value of the expression is provided in the answer. This, of course, was not required of the participants.Comment 2. The answer can be simplified to one inverse trigonometric function. Indeed, if alpha=operatorname{arctg} frac{2}{9}, beta=operatorname{arctg} frac{6}{7}, thenoperatorname{tg}(alpha+beta)=frac{operatorname{tg} alpha+operatorname{tg} beta}{1-operatorname{tg} alpha operatorname{tg} beta}=frac{2 / 9+6 / 7}{1-2 / 9 cdot 6 / 7}=frac{4}{3}from which, since alpha+beta inleft(-frac{pi}{2} ; frac{pi}{2}right), we get that the answer is frac{1}{8} operatorname{arctg} frac{4}{3}. Such a simplification was not required of the participants.Solution. Divide both sides of the equation by sqrt{85}=sqrt{2^{2}+9^{2}}=sqrt{6^{2}+7^{2}}. Let alpha=operatorname{arctg} frac{2}{9}, beta=operatorname{arctg} frac{6}{7}; then sin alpha=frac{2}{sqrt{85}}, cos alpha=frac{9}{sqrt{85}}, sin beta=frac{6}{sqrt{85}}, cos alpha=frac{7}{sqrt{85}}, and the equation takes the formsin alpha sin 6 x+cos alpha cos 6 x=sin beta sin 2 x+cos beta cos 2 xor cos (6 x-alpha)=cos (2 x-beta). The solutions to this equation are the setleft[begin{array}{l}6 x-alpha=2 x-beta+2 pi k, k in mathbb{Z} 6 x-alpha=beta-2 x+2 pi n, n in mathbb{Z}end{array}right.that is,left[begin{array}{l}x=frac{alpha-beta+2 pi k}{4}, k in mathbb{Z} x=frac{alpha+beta+2 pi n}{8}, n in mathbb{Z}end{array}right.Since 0<alpha<beta<frac{pi}{2}, the smallest positive roots of the series in the set are the numbers x_{1}=frac{alpha-beta+2 pi}{4} and x_{2}=frac{alpha+beta}{8}. It remains to compare x_{1} and x_{2}. For this, note thatx_{2}=frac{alpha+beta}{8}<frac{pi}{8}<frac{pi}{4}<frac{2 pi+alpha-beta}{4}=x_{1}

🧠:[<think>Okay, so Zhenya has these cards numbered from 1 to 9, but he lost the card with number 7. Now, the question is whether he can arrange the remaining 8 cards in a row such that any two adjacent cards form a number divisible by 7. Hmm, interesting. Let me try to figure this out step by step.First, let's confirm which cards are left. The original set is 1,2,3,4,5,6,7,8,9. Since he lost the 7, the remaining cards are 1,2,3,4,5,6,8,9. So we need to arrange these 8 cards in a sequence where each pair of neighboring cards forms a two-digit number divisible by 7. Alright, so the two-digit numbers we can form with adjacent cards must be divisible by 7. Let me list all possible two-digit numbers from the remaining cards and check which ones are divisible by 7.The remaining digits are 1,2,3,4,5,6,8,9. Let's pair them up in all possible ways (without repeating digits) and check divisibility by 7. But wait, since we're arranging them in a row, each digit (except the first and last) will be part of two two-digit numbers: one as the tens place with the previous digit and one as the units place with the next digit. But maybe first, just list all possible two-digit numbers from these digits and see which ones are divisible by 7.Possible two-digit numbers (tens and units digits must be different, and 7 is excluded):12,13,14,15,16,18,19,21,23,24,25,26,28,29,31,32,34,35,36,38,39,41,42,43,45,46,48,49,51,52,53,54,56,58,59,61,62,63,64,65,68,69,81,82,83,84,85,86,89,91,92,93,94,95,96,98.Now, check which of these are divisible by 7. Let me go through each number:12 ÷ 7 = 1.714... → No13 ÷ 7 ≈ 1.857... → No14 ÷ 7 = 2 → Yes15 ÷ 7 ≈ 2.142... → No16 ÷ 7 ≈ 2.285... → No18 ÷ 7 ≈ 2.571... → No19 ÷ 7 ≈ 2.714... → No21 ÷ 7 = 3 → Yes23 ÷ 7 ≈ 3.285... → No24 ÷ 7 ≈ 3.428... → No25 ÷ 7 ≈ 3.571... → No26 ÷ 7 ≈ 3.714... → No28 ÷ 7 = 4 → Yes29 ÷ 7 ≈ 4.142... → No31 ÷ 7 ≈ 4.428... → No32 ÷ 7 ≈ 4.571... → No34 ÷ 7 ≈ 4.857... → No35 ÷ 7 = 5 → Yes36 ÷ 7 ≈ 5.142... → No38 ÷ 7 ≈ 5.428... → No39 ÷ 7 ≈ 5.571... → No41 ÷ 7 ≈ 5.857... → No42 ÷ 7 = 6 → Yes43 ÷ 7 ≈ 6.142... → No45 ÷ 7 ≈ 6.428... → No46 ÷ 7 ≈ 6.571... → No48 ÷ 7 ≈ 6.857... → No49 ÷ 7 = 7 → Yes, but 49 uses the digit 4 and 9. But wait, 49 is 7 squared, but the card with 7 is lost. Wait, hold on, 49 is made by 4 and 9, so since 7 is lost, but 4 and 9 are still present. So 49 is allowed. However, 49 ÷ 7 = 7, which is exact. So 49 is allowed. But we need to check if 49 is formed by adjacent cards. But since 7 is lost, but the numbers 4 and 9 can be adjacent, so 49 is okay.51 ÷ 7 ≈ 7.285... → No52 ÷ 7 ≈ 7.428... → No53 ÷ 7 ≈ 7.571... → No54 ÷ 7 ≈ 7.714... → No56 ÷ 7 = 8 → Yes58 ÷ 7 ≈ 8.285... → No59 ÷ 7 ≈ 8.428... → No61 ÷ 7 ≈ 8.714... → No62 ÷ 7 ≈ 8.857... → No63 ÷ 7 = 9 → Yes, but wait, 63 ÷ 7 = 9. But 63 is 6 and 3. 6 and 3 are both available. So 63 is okay.64 ÷ 7 ≈ 9.142... → No65 ÷ 7 ≈ 9.285... → No68 ÷ 7 ≈ 9.714... → No69 ÷ 7 ≈ 9.857... → No81 ÷ 7 ≈ 11.571... → No82 ÷ 7 ≈ 11.714... → No83 ÷ 7 ≈ 11.857... → No84 ÷ 7 = 12 → Yes85 ÷ 7 ≈ 12.142... → No86 ÷ 7 ≈ 12.285... → No89 ÷ 7 ≈ 12.714... → No91 ÷ 7 = 13 → Yes92 ÷ 7 ≈ 13.142... → No93 ÷ 7 ≈ 13.285... → No94 ÷ 7 ≈ 13.428... → No95 ÷ 7 ≈ 13.571... → No96 ÷ 7 ≈ 13.714... → No98 ÷ 7 = 14 → YesSo compiling all the two-digit numbers divisible by 7 that can be formed with the remaining digits (excluding 7):14, 21, 28, 35, 42, 49, 56, 63, 84, 91, 98.Let me list them again:- 14 (1 and 4)- 21 (2 and 1)- 28 (2 and 8)- 35 (3 and 5)- 42 (4 and 2)- 49 (4 and 9)- 56 (5 and 6)- 63 (6 and 3)- 84 (8 and 4)- 91 (9 and 1)- 98 (9 and 8)So these are the possible transitions between the digits. Each of these two-digit numbers corresponds to a pair of digits that can be adjacent in the sequence. Our goal is to arrange the 8 digits (1,2,3,4,5,6,8,9) in a sequence where each adjacent pair is one of these numbers. Importantly, each digit can be used only once because we have only one of each card.This problem is similar to constructing a path in a graph where each node is a digit, and edges represent the allowable transitions (i.e., the two-digit number is divisible by 7). So, we need to find a Hamiltonian path in this graph— visiting each node exactly once.Let me try to model this as a graph. The nodes are digits 1,2,3,4,5,6,8,9. The edges are based on the two-digit numbers above.From the list:14: 1 -> 421: 2 -> 128: 2 -> 835: 3 -> 542: 4 -> 249: 4 -> 956: 5 -> 663: 6 -> 384: 8 -> 491: 9 -> 198: 9 -> 8So, let's represent this as directed edges:1 has outgoing edges to 4 (from 14).2 has outgoing edges to 1 (21) and 8 (28).3 has outgoing edge to 5 (35).4 has outgoing edges to 2 (42) and 9 (49).5 has outgoing edge to 6 (56).6 has outgoing edge to 3 (63).8 has outgoing edge to 4 (84).9 has outgoing edges to 1 (91) and 8 (98).Now, each node's in-degree and out-degree:Let's compute in-degrees and out-degrees:1:- In-degrees: from 2 (21), from 9 (91) → in-degree 2- Out-degrees: to 4 (14) → out-degree 12:- In-degrees: from 4 (42) → in-degree 1- Out-degrees: to 1,8 (21, 28) → out-degree 23:- In-degrees: from 6 (63) → in-degree 1- Out-degrees: to 5 (35) → out-degree 14:- In-degrees: from 1 (14), from 8 (84) → in-degree 2- Out-degrees: to 2,9 (42,49) → out-degree 25:- In-degrees: from 3 (35) → in-degree 1- Out-degrees: to 6 (56) → out-degree 16:- In-degrees: from 5 (56) → in-degree 1- Out-degrees: to 3 (63) → out-degree 18:- In-degrees: from 2 (28), from 9 (98) → in-degree 2- Out-degrees: to 4 (84) → out-degree 19:- In-degrees: from 4 (49) → in-degree 1- Out-degrees: to 1,8 (91,98) → out-degree 2In a directed graph, for a Hamiltonian path to exist, it should satisfy the conditions for a directed Hamiltonian path. Generally, for a directed graph, a Hamiltonian path exists if the graph is strongly connected, and for all vertices except two, the in-degree equals the out-degree, and for the two exceptional vertices, one has in-degree one more than out-degree (end vertex), and the other has out-degree one more than in-degree (start vertex). However, since the graph might not be strongly connected, this is just a general condition, not a guarantee.But let's check in-degrees and out-degrees for each node:Node | In-degree | Out-degree1 | 2 | 12 | 1 | 23 | 1 | 14 | 2 | 25 | 1 | 16 | 1 | 18 | 2 | 19 | 1 | 2So, nodes where in-degree ≠ out-degree:Node 1: in - out = 2 - 1 = +1Node 2: in - out = 1 - 2 = -1Node 8: in - out = 2 - 1 = +1Node 9: in - out = 1 - 2 = -1So we have two nodes (1 and 8) with in-degree exceeding out-degree by 1, and two nodes (2 and 9) with out-degree exceeding in-degree by 1. For a directed Eulerian trail (which is a trail that visits every edge exactly once), we need exactly two nodes with in-degree = out-degree +1 (end nodes) and two nodes with out-degree = in-degree +1 (start nodes). However, in our case, we are not looking for an Eulerian trail but a Hamiltonian path, which is a path that visits every node exactly once. The two concepts are related but different.In a Hamiltonian path in a directed graph, typically, all nodes except the start and end have equal in-degree and out-degree, but this is not a strict rule because the graph might not be strongly connected. However, in our case, since there are multiple nodes with imbalanced degrees, it complicates things.Alternatively, maybe we can model this as an undirected graph and check for Hamiltonian path. But even then, we have to consider the directionality because the two-digit number must be formed in a specific order. For example, 14 is different from 41; 14 is allowed (divisible by 7) but 41 isn't. So direction matters here.Therefore, the problem reduces to finding a directed Hamiltonian path in this graph. Let's see if such a path exists.One approach is to try to construct such a path manually, starting from nodes that have more out-degrees than in-degrees (potential start nodes) and ending at nodes with more in-degrees than out-degrees (potential end nodes). From our earlier analysis, nodes 2 and 9 have out-degrees exceeding in-degrees, so they could be start nodes. Nodes 1 and 8 have in-degrees exceeding out-degrees, so they could be end nodes.So possible paths could start at 2 or 9 and end at 1 or 8.Let me try starting at 2.Starting from 2:From node 2, we can go to 1 or 8.First, let's try 2 -> 1.Then, from 1, the only outgoing edge is to 4.So path: 2 -> 1 -> 4.From 4, we can go to 2 or 9. But 2 is already used, so go to 9.Path: 2 ->1 ->4 ->9.From 9, go to 8 or 1. 1 is already used, so go to 8.Path: 2->1->4->9->8.From 8, go to 4. But 4 is already used. Dead end. So this path ends here with only 5 nodes. Not enough. Let's backtrack.Alternative from 4: instead of going to 9, can we go to 2? But 2 is already used. So no. So this path is stuck.Alternative, starting at 2 ->8.Path: 2->8.From 8, go to 4.Path: 2->8->4.From 4, go to 2 or 9. 2 is used, so 9.Path: 2->8->4->9.From 9, go to 1 or 8. 8 is used, so 1.Path: 2->8->4->9->1.From 1, go to 4. Already used. Dead end. Only 5 nodes. Hmm.Alternatively, starting at 2, maybe a different path.Wait, let's try starting at 9.Starting node 9.From 9, can go to 1 or 8.First, 9->1.Path: 9->1.From 1, go to 4.Path: 9->1->4.From 4, go to 2 or 9. 9 is used, so 2.Path:9->1->4->2.From 2, go to 1 or 8. 1 is used, so 8.Path:9->1->4->2->8.From 8, go to 4. Used. Dead end. 5 nodes. Not good.Alternative, starting at 9->8.Path:9->8.From 8, go to 4.Path:9->8->4.From 4, go to 2 or 9. 9 is used, so 2.Path:9->8->4->2.From 2, go to 1 or 8. 8 is used, so 1.Path:9->8->4->2->1.From 1, go to 4. Used. Dead end. Again 5 nodes.Hmm. Neither starting at 2 nor 9 gives a full path. Let's try another approach.Maybe the path isn't starting at 2 or 9. Wait, but according to the degrees, the start should be 2 or 9. But perhaps the graph isn't connected in a way that allows a Hamiltonian path from those. Maybe we need to consider cycles?Wait, maybe there's a cycle in the graph. Let's check.Looking at the edges:1 ->44->2, 4->92->1, 2->88->49->1, 9->8So the subgraph involving 1,2,4,8,9 forms a cycle? Let's see:For example, 1->4->2->8->4... Wait, 1->4->2->1 would form a cycle, but 1->4->2->8->4 is a cycle from 4->2->8->4. Similarly, 9 connects to 1 and 8.So maybe there are cycles here. If there's a cycle, but we need to traverse all 8 nodes. So perhaps combining cycles with other edges.Looking at the other nodes: 3,5,6.Edges for these:3->55->66->3So 3->5->6->3 is a cycle.Therefore, the graph has two separate cycles: one involving 3,5,6 and another involving 1,2,4,8,9. But wait, is that true? Wait, is there a connection between these two cycles?Looking at the edges, node 4 connects to 2 and 9; node 9 connects to 1 and 8; node 8 connects to 4. The other cycle is 3->5->6->3. There are no edges connecting these two cycles. So the graph is disconnected into two components: one with 1,2,4,8,9 and another with 3,5,6.Therefore, it's impossible to have a Hamiltonian path that connects all 8 nodes, since the graph is disconnected into two separate components. So you can't traverse from one component to the other. Hence, arranging all 8 cards in a row is impossible.Wait, but that seems like a conclusion. Let me confirm.So, if the graph is split into two disconnected components, one with 1,2,4,8,9 (5 nodes) and another with 3,5,6 (3 nodes), then a Hamiltonian path would need to go from one component to the other, which is impossible because there are no edges between them. Hence, you can't have a sequence that uses all 8 digits, as moving from one component to the other isn't possible.Therefore, the answer is no, it's not possible to arrange the remaining 8 cards in such a way.Wait, but before finalizing that conclusion, let me double-check. Is the graph indeed disconnected?Nodes 1,2,4,8,9 are connected among themselves, as per the edges:1->4, 4->2, 4->9, 2->1, 2->8, 8->4, 9->1, 9->8.Yes, they form a strongly connected component? Let's see. For example, can we get from any node in this set to any other?From 1: 1->4, then 4 can go to 2 or 9. From 2, you can go to 1 or 8. From 8, go to 4. From 4, go to 2 or 9. From 9, go to 1 or 8. So yes, within this group, you can reach any node from any other node. So it's a strongly connected component.The other component is 3,5,6:3->5, 5->6, 6->3. So it's a cycle. Also strongly connected.Therefore, the graph has two strongly connected components with no edges between them. Therefore, any path must stay within a component. Since one component has 5 nodes and the other 3, we cannot have a path that includes all 8 nodes. Therefore, it's impossible.Hence, the answer is no, such an arrangement is not possible.Final Answerboxed{text{No}}</think>Zhenya has 8 cards remaining: 1, 2, 3, 4, 5, 6, 8, 9. We need to determine if these can be arranged in a row such that each pair of adjacent cards forms a number divisible by 7.First, we list all possible two-digit numbers formed by these digits and check their divisibility by 7. The valid numbers are: 14, 21, 28, 35, 42, 49, 56, 63, 84, 91, 98. We model this as a directed graph where each node represents a digit, and edges represent valid transitions. The graph has the following edges:- 1 → 4- 2 → 1, 2 → 8- 3 → 5- 4 → 2, 4 → 9- 5 → 6- 6 → 3- 8 → 4- 9 → 1, 9 → 8Next, we analyze the in-degrees and out-degrees of each node:- Node 1: in-degree 2, out-degree 1- Node 2: in-degree 1, out-degree 2- Node 3: in-degree 1, out-degree 1- Node 4: in-degree 2, out-degree 2- Node 5: in-degree 1, out-degree 1- Node 6: in-degree 1, out-degree 1- Node 8: in-degree 2, out-degree 1- Node 9: in-degree 1, out-degree 2We observe that the graph is divided into two strongly connected components:1. Nodes 1, 2, 4, 8, 9 form a cycle.2. Nodes 3, 5, 6 form another cycle.Since there are no edges connecting these two components, it is impossible to traverse all 8 nodes in a single path. Therefore, arranging all 8 cards to meet the given condition is not possible.boxed{text{No}},<think>Okay, so Zhenya has these cards numbered from 1 to 9, but he lost the card with number 7. Now he wants to arrange the remaining 8 cards in a row such that any two adjacent cards form a number divisible by 7. Hmm, interesting problem. Let me think through this step by step.First, let's list out the cards he has left. The original set is 1 through 9, so without the 7, the remaining cards are 1, 2, 3, 4, 5, 6, 8, 9. That's eight cards. The task is to arrange them in a sequence where each pair of adjacent cards forms a two-digit number divisible by 7. So, for example, if we have cards 1 and 4 next to each other, that would form 14 or 41. We need to check if either 14 or 41 is divisible by 7. Since 14 ÷ 7 = 2, that works. But 41 ÷ 7 is about 5.857, which is not an integer. So in that case, the order would matter: 1 followed by 4 is okay, but 4 followed by 1 is not.Therefore, for each pair of adjacent cards, the two-digit number formed must be divisible by 7, considering the order. So maybe I need to figure out all possible two-digit combinations from the remaining cards and see which ones are divisible by 7. Then, model this as a graph where each node is a card, and edges represent allowed transitions (i.e., two cards can be adjacent if their two-digit number is divisible by 7). Then, the problem reduces to finding a Hamiltonian path in this graph.Let me start by listing all possible two-digit numbers from the remaining cards and check their divisibility by 7.The remaining digits are 1, 2, 3, 4, 5, 6, 8, 9. Let's list all possible two-digit combinations (both forward and reverse):12, 21, 13, 31, 14, 41, 15, 51, 16, 61, 18, 81, 19, 91,23, 32, 24, 42, 25, 52, 26, 62, 28, 82, 29, 92,34, 43, 35, 53, 36, 63, 38, 83, 39, 93,45, 54, 46, 64, 48, 84, 49, 94,56, 65, 58, 85, 59, 95,68, 86, 69, 96,89, 98.Now, check which of these are divisible by 7.Let's divide each by 7:12 ÷ 7 = 1.714… No.21 ÷ 7 = 3. Exactly, so 21 is divisible by 7.13 ÷ 7 = 1.857… No.31 ÷ 7 ≈ 4.428… No.14 ÷ 7 = 2. Exactly, yes.41 ÷ 7 ≈ 5.857… No.15 ÷ 7 ≈ 2.142… No.51 ÷ 7 ≈ 7.285… No.16 ÷ 7 ≈ 2.285… No.61 ÷ 7 ≈ 8.714… No.18 ÷ 7 ≈ 2.571… No.81 ÷ 7 ≈ 11.571… No.19 ÷ 7 ≈ 2.714… No.91 ÷ 7 = 13. Exactly, yes.23 ÷ 7 ≈ 3.285… No.32 ÷ 7 ≈ 4.571… No.24 ÷ 7 ≈ 3.428… No.42 ÷ 7 = 6. Exactly, yes.25 ÷ 7 ≈ 3.571… No.52 ÷ 7 ≈ 7.428… No.26 ÷ 7 ≈ 3.714… No.62 ÷ 7 ≈ 8.857… No.28 ÷ 7 = 4. Exactly, yes.82 ÷ 7 ≈ 11.714… No.29 ÷ 7 ≈ 4.142… No.92 ÷ 7 ≈ 13.142… No.34 ÷ 7 ≈ 4.857… No.43 ÷ 7 ≈ 6.142… No.35 ÷ 7 = 5. Exactly, yes.53 ÷ 7 ≈ 7.571… No.36 ÷ 7 ≈ 5.142… No.63 ÷ 7 = 9. Exactly, yes.38 ÷ 7 ≈ 5.428… No.83 ÷ 7 ≈ 11.857… No.39 ÷ 7 ≈ 5.571… No.93 ÷ 7 ≈ 13.285… No.45 ÷ 7 ≈ 6.428… No.54 ÷ 7 ≈ 7.714… No.46 ÷ 7 ≈ 6.571… No.64 ÷ 7 ≈ 9.142… No.48 ÷ 7 ≈ 6.857… No.84 ÷ 7 = 12. Exactly, yes.49 ÷ 7 = 7. Exactly, yes.94 ÷ 7 ≈ 13.428… No.56 ÷ 7 = 8. Exactly, yes.65 ÷ 7 ≈ 9.285… No.58 ÷ 7 ≈ 8.285… No.85 ÷ 7 ≈ 12.142… No.59 ÷ 7 ≈ 8.428… No.95 ÷ 7 ≈ 13.571… No.68 ÷ 7 ≈ 9.714… No.86 ÷ 7 ≈ 12.285… No.69 ÷ 7 ≈ 9.857… No.96 ÷ 7 ≈ 13.714… No.89 ÷ 7 ≈ 12.714… No.98 ÷ 7 = 14. Exactly, yes.So, the two-digit numbers divisible by 7 from these digits are:21, 14, 91, 42, 28, 35, 63, 84, 49, 56, 98.So each of these corresponds to an edge in the graph. Let's note them down as directed edges (since order matters):From 2 to 1 (21),From 1 to 4 (14),From 9 to 1 (91),From 4 to 2 (42),From 2 to 8 (28),From 3 to 5 (35),From 6 to 3 (63),From 8 to 4 (84),From 4 to 9 (49),From 5 to 6 (56),From 9 to 8 (98).Wait, let me verify each of these:- 21: 2 → 1- 14: 1 → 4- 91: 9 → 1- 42: 4 → 2- 28: 2 → 8- 35: 3 → 5- 63: 6 → 3- 84: 8 → 4- 49: 4 → 9- 56: 5 → 6- 98: 9 → 8So these are the directed edges. Now, our graph has nodes 1,2,3,4,5,6,8,9, and the edges as above. The question is: is there a Hamiltonian path in this directed graph? That is, a path that visits each node exactly once, following the direction of the edges.To check this, we can try to model the graph and look for such a path. Alternatively, we can analyze the structure.First, let's note the in-degrees and out-degrees of each node. For a directed graph to have a Hamiltonian path, it's necessary (but not sufficient) that the graph is strongly connected, or at least connected in the underlying undirected graph. Also, typically, for a Hamiltonian path in a directed graph (also called a directed path), we can have at most one node with in-degree 0 (start node) and at most one node with out-degree 0 (end node). Let's check the degrees.Calculating in-degrees and out-degrees:Node 1:- In-degrees: edges coming into 1: 2 → 1 (from 2), 9 → 1 (from 9). So in-degree 2.- Out-degrees: edges going out from 1: 1 → 4. So out-degree 1.Node 2:- In-degrees: 4 → 2 (from 4). So in-degree 1.- Out-degrees: 2 → 1, 2 → 8. So out-degree 2.Node 3:- In-degrees: 6 → 3 (from 6). So in-degree 1.- Out-degrees: 3 → 5. So out-degree 1.Node 4:- In-degrees: 1 → 4 (from 1), 8 → 4 (from 8). So in-degree 2.- Out-degrees: 4 → 2, 4 → 9. So out-degree 2.Node 5:- In-degrees: 3 → 5 (from 3). So in-degree 1.- Out-degrees: 5 → 6. So out-degree 1.Node 6:- In-degrees: 5 → 6 (from 5). So in-degree 1.- Out-degrees: 6 → 3. So out-degree 1.Node 8:- In-degrees: 2 → 8 (from 2), 9 → 8 (from 9). So in-degree 2.- Out-degrees: 8 → 4. So out-degree 1.Node 9:- In-degrees: 4 → 9 (from 4). So in-degree 1.- Out-degrees: 9 → 1, 9 → 8. So out-degree 2.So, summarizing:Node | In-degree | Out-degree1 | 2 | 12 | 1 | 23 | 1 | 14 | 2 | 25 | 1 | 16 | 1 | 18 | 2 | 19 | 1 | 2For a directed Hamiltonian path, there should be one node with out-degree = in-degree + 1 (start node) and one node with in-degree = out-degree + 1 (end node). All other nodes must have equal in-degree and out-degree.Looking at the degrees:Let's check the differences (out - in):1: 1 - 2 = -12: 2 - 1 = +13: 1 - 1 = 04: 2 - 2 = 05: 1 - 1 = 06: 1 - 1 = 08: 1 - 2 = -19: 2 - 1 = +1So, nodes 2 and 9 have out-degree exceeding in-degree by 1, and nodes 1 and 8 have in-degree exceeding out-degree by 1. In a directed Eulerian trail (which is different from a Hamiltonian path), we have such conditions, but here we're looking for a path that visits each node exactly once, not each edge. However, similar principles might apply. Wait, perhaps not directly.Alternatively, in a Hamiltonian path in a directed graph, the start node can have out-degree one more than in-degree, and the end node can have in-degree one more than out-degree. However, here we have two nodes (2 and 9) with out-degree exceeding in-degree by 1, and two nodes (1 and 8) with in-degree exceeding out-degree by 1. Since a Hamiltonian path can have only one start and one end, this suggests that perhaps such a path is not possible because we have two potential start nodes and two potential end nodes.But this is a necessary condition for an Eulerian trail, not necessarily for a Hamiltonian path. The Eulerian trail conditions are about edge traversal, whereas Hamiltonian path is about node traversal. So maybe the in-degree and out-degree aren't directly applicable here. Hmm. Maybe that line of thought isn't helpful here.Alternatively, perhaps we can attempt to construct the path manually.Let me try to build the path step by step.Looking at the edges, let's see possible starting points. If we start with a node that has no incoming edges, but in our case, all nodes have incoming edges except maybe none. Wait, nodes 1, 2, 3, 4, 5, 6, 8, 9 all have in-degrees at least 1 except... Wait, no, all nodes have in-degree at least 1:Node 1: in-degree 2Node 2: in-degree 1Node 3: in-degree 1Node 4: in-degree 2Node 5: in-degree 1Node 6: in-degree 1Node 8: in-degree 2Node 9: in-degree 1So every node has at least one incoming edge, which complicates starting a path, because theoretically, you could start anywhere, but you have to make sure that you can enter every other node.Alternatively, since it's a Hamiltonian path, which is a sequence visiting each node once, maybe starting at a node with high out-degree. For example, nodes 2, 4, and 9 have out-degree 2.Let's consider starting at node 2. From node 2, we can go to 1 or 8.Option 1: Start at 2 → 1. Then from 1, we can only go to 4. Then from 4, we can go to 2 or 9. But 2 is already visited, so go to 9. From 9, we can go to 1 or 8. 1 is already visited, so go to 8. From 8, go to 4. But 4 is already visited. Dead end. So this path is 2→1→4→9→8. Then stuck. Only visited 5 nodes. Not all 8.Option 2: Start at 2 → 8. Then from 8 → 4. From 4, can go to 2 or 9. 2 is already visited, so go to 9. From 9, go to 1 or 8. 8 is visited, so go to 1. From 1 → 4, but 4 is visited. Dead end. So path: 2→8→4→9→1. Stuck at 1, visited 5 nodes.Alternatively, maybe starting elsewhere. Let's try starting at 9. From 9, can go to 1 or 8.Option 1: Start at 9 →1. Then 1→4. From 4, go to 2 or 9 (9 is visited). So 4→2. From 2, go to 1 or 8 (1 is visited). So 2→8. From 8→4 (visited). Stuck. Path: 9→1→4→2→8. Only 5 nodes.Option 2: Start at 9→8. Then 8→4. From 4, go to 2 or 9. 9 is visited. So 4→2. From 2, go to 1 or 8 (8 is visited). So 2→1. From 1→4 (visited). Stuck. Path: 9→8→4→2→1. Again 5 nodes.Alternatively, starting at 4. From 4, go to 2 or 9.Start 4→2. Then 2→1 or 8. Let's say 2→1. Then 1→4 (visited). Dead end. If 2→8, then 8→4 (visited). So no good.Start 4→9. Then 9→1 or 8. If 9→1, then 1→4 (visited). If 9→8, then 8→4 (visited). Also no.Hmm. Starting with nodes that have high out-degree doesn't seem to help. Let's try starting with node 3. From 3, we can only go to 5. Then from 5→6. From 6→3 (already visited). So path: 3→5→6. Only 3 nodes.Alternatively, start with node 5. 5→6→3→5? No, same issue.Start with node 6. 6→3→5→6. Again loop.Node 7 is missing, so 7 isn't here. Let's see node 8. From 8→4, but starting at 8, then 8→4. Then from 4→2 or 9. Then maybe 4→2. From 2→1 or 8. 8 is visited, so 2→1. Then 1→4 (visited). So path: 8→4→2→1. Then stuck.Alternatively, 8→4→9. From 9→1 or 8. 8 is visited, so 9→1. Then 1→4 (visited). So same as before.Hmm. Maybe we need to connect different segments.Looking at the edges, there are a few separate components? Let me check the connectivity.From the edges:- 2 connects to 1 and 8.- 1 connects to 4.- 4 connects to 2 and 9.- 9 connects to 1 and 8.- 8 connects to 4.So nodes 1,2,4,8,9 are interconnected.Then node 3 connects to 5.Node 5 connects to 6.Node 6 connects to 3.So nodes 3,5,6 form a cycle: 3→5→6→3.Additionally, node 8 connects to 4, which is in the first group. Wait, but node 8 is in the first group. Wait, maybe the entire graph is connected?Wait, from the first group (1,2,4,8,9), node 8 is connected to 4, which is connected to 2 and 9, etc. Then the other group is 3,5,6. Are they connected to the main group? Let's see. The edges:Is there any connection between the first group (1,2,4,8,9) and the second group (3,5,6)? Looking at the edges:From the first group, nodes are 1,2,4,8,9. The other nodes are 3,5,6. Are there any edges from the first group to the second or vice versa?Looking at our edge list:Edges are:2→1, 2→8,1→4,9→1, 9→8,4→2, 4→9,8→4,3→5,5→6,6→3.So, no edges connecting the first group (1,2,4,8,9) with the second group (3,5,6). Therefore, the graph is split into two separate components: the first component with nodes 1,2,4,8,9 and the second component with nodes 3,5,6. Therefore, a Hamiltonian path that includes all 8 nodes is impossible, because the graph is disconnected into two components. You can't traverse from one component to the other, as there are no edges between them. Therefore, it's impossible to arrange the cards in such a way.Wait a minute, hold on. Let me verify this. The problem states that Zhenya has 8 cards: 1,2,3,4,5,6,8,9. The two components in the graph are {1,2,4,8,9} and {3,5,6}. These are 5 and 3 nodes respectively. But we need a single path covering all 8 nodes. Since the graph is disconnected into two separate components with no edges between them, it's impossible to have a path that goes through all 8 nodes. Therefore, the answer should be no, it's not possible.But let me just make sure I didn't miss any edges between the components. Let me double-check the list of edges. The edges are:From 2 to 1, 2 to 8,1 to 4,9 to 1, 9 to 8,4 to 2, 4 to 9,8 to 4,3 to 5,5 to 6,6 to 3.No edges connecting, say, 6 to any node in the first group, or 3 to any node in the first group, etc. Therefore, indeed, the graph is disconnected. Hence, a Hamiltonian path cannot exist because you can't move from one component to the other.Therefore, the conclusion is that it's impossible to arrange the remaining 8 cards in such a way.But wait, let me check once more if I missed any edges. For example, is there a two-digit number between the two components that I might have overlooked?For example, from node 3 to node 1: 31 is not divisible by 7. 13 is not. From node 5 to node 1: 51 isn't. 15 isn't. From node 6 to node 1: 61 isn't. 16 isn't. From node 3 to node 2: 32 isn't. 23 isn't. From node 3 to 4: 34 isn't, 43 isn't. From node 5 to 2: 52 isn't, 25 isn't. From node 5 to 4: 54 isn't, 45 isn't. From node 6 to 2: 62 isn't, 26 isn't. From node 6 to 4: 64 isn't, 46 isn't. Similarly, from first group to the second group: 1 to 3: 13, 31; neither divisible by 7. 1 to 5: 15, 51; nope. 1 to 6:16,61; nope. 2 to 3:23,32; nope. 2 to5:25,52; nope. 2 to6:26,62; nope. 4 to3:43,34; nope. 4 to5:45,54; nope. 4 to6:46,64; nope. 8 to3:83,38; nope. 8 to5:85,58; nope. 8 to6:86,68; nope. 9 to3:93,39; nope. 9 to5:95,59; nope. 9 to6:96,69; nope. So indeed, there are no edges connecting the two components.Therefore, since the graph is disconnected, a Hamiltonian path covering all 8 nodes is impossible. Therefore, the answer is no, it cannot be arranged.But wait, the problem says "arrange the remaining 8 cards in a row". Since all the numbers must be adjacent, but maybe the two components can be connected in some way? Wait, but if there are no edges between the components, how can you transition from one to the other? You can't. So unless there's a mistake in my edge list.Wait, let's confirm again the two-digit numbers divisible by 7. Maybe I missed some.Earlier, I listed:21, 14, 91, 42, 28, 35, 63, 84, 49, 56, 98.Wait, 84 is 12x7, which is 84. 49 is 7x7, but 7 is excluded. Wait, 49 is allowed because 4 and 9 are still present. 56 is 7x8. Yes. 98 is 14x7. So all those are correct.Is there any other two-digit number using the remaining digits that I missed?Let me check 89 and 98: 98 is divisible by 7 (14x7), but 89 is 12.714..., so not. 89 is not. What about 35: 35 is 5x7. Correct. 63 is 9x7. Correct. 42 is 6x7. Correct. 21 is 3x7. Correct. 14 is 2x7. Correct. 28 is 4x7. Correct. 56 is 8x7. Correct. 84 is 12x7. Correct. 49 is 7x7. But since 7 is missing, but 4 and 9 are present, so 49 is formed by 4 and 9. Correct. 91 is 13x7. Correct. 98 is 14x7. Correct. So those are all the two-digit numbers possible with the remaining digits. So the edge list is accurate.Therefore, the graph is indeed split into two disconnected components. Therefore, arranging all 8 cards in a row with adjacent numbers divisible by 7 is impossible. Hence, the answer is no.But wait, the problem is in Russian, and the name is Zhenya, which is a Russian diminutive for Evgeny. Just to make sure, did I interpret the problem correctly? The problem states: "Can the remaining 8 cards be arranged in a row so that any two adjacent cards form a number divisible by 7?" So adjacent cards, meaning that if you have card A followed by card B, the number AB must be divisible by 7. Not BA. So the direction matters, which we considered by making directed edges. And the conclusion is that since the graph is disconnected into two components, you cannot have a single path covering all nodes. Therefore, the answer is no.However, just to be thorough, let's check if there's a possibility that maybe starting in one component and somehow jumping to the other, but given the constraints, since you can only move along directed edges, and there are no edges between components, it's impossible. Therefore, the answer is no.Final Answerboxed{text{No}}</think>Zhenya has 8 cards left: 1, 2, 3, 4, 5, 6, 8, 9. We need to determine if these cards can be arranged in a row such that any two adjacent cards form a number divisible by 7.First, we list all two-digit numbers formed by the remaining cards and check their divisibility by 7. The valid two-digit numbers and their corresponding edges are:- 21 (2 → 1)- 14 (1 → 4)- 91 (9 → 1)- 42 (4 → 2)- 28 (2 → 8)- 35 (3 → 5)- 63 (6 → 3)- 84 (8 → 4)- 49 (4 → 9)- 56 (5 → 6)- 98 (9 → 8)Next, we analyze the graph structure. The graph has two disconnected components:1. Component 1: Nodes 1, 2, 4, 8, 9 with edges among them.2. Component 2: Nodes 3, 5, 6 with edges forming a cycle (3 → 5 → 6 → 3).Since there are no edges connecting the two components, it is impossible to form a single path that visits all nodes. A Hamiltonian path covering all 8 nodes is not possible in a disconnected graph.Thus, the remaining 8 cards cannot be arranged in a row to satisfy the given condition.[boxed{text{No}}]]

🔑:3. Answer: no.Note that after 5 there must be 6, after 6-3, after 3-5, after 5-6. No other digits can precede these either. Therefore, it is impossible to arrange all these digits in one row.

🧠:[<think>Okay, let's see. I need to find the common difference d of an arithmetic sequence given that the difference between the average of the first 2017 terms and the average of the first 17 terms is 100. The options are given, so maybe after solving, I can check which one fits. Let me start by recalling some formulas related to arithmetic sequences.First, the sum of the first n terms of an arithmetic sequence is given by S_n = (n/2)[2a_1 + (n - 1)d], where a_1 is the first term and d is the common difference. Alternatively, this can also be written as S_n = n(a_1 + a_n)/2, since a_n = a_1 + (n - 1)d. Either formula should work, but I'll go with the first one because it directly includes d.The problem mentions the average of the first n terms, which is S_n divided by n. So, S_n / n = [ (n/2)(2a_1 + (n - 1)d ) ] / n = (2a_1 + (n - 1)d)/2. Therefore, the average of the first n terms simplifies to (a_1 + a_n)/2, which makes sense because in an arithmetic sequence, the average is just the average of the first and last term.Given that, the problem states that S_{2017}/2017 - S_{17}/17 = 100. Let me write that out using the average formula:( (2a_1 + (2017 - 1)d ) / 2 ) - ( (2a_1 + (17 - 1)d ) / 2 ) = 100Simplify inside the parentheses:(2a1 + 2016d)/2 - (2a1 + 16d)/2 = 100Now, let's compute each term:First term: (2a1 + 2016d)/2 = a1 + 1008dSecond term: (2a1 + 16d)/2 = a1 + 8dSubtracting the second term from the first:(a1 + 1008d) - (a1 + 8d) = 100Simplify:a1 - a1 + 1008d - 8d = 100So, (1008d - 8d) = 1000d = 100Therefore, 1000d = 100 => d = 100 / 1000 = 1/10Wait, so d is 1/10? Let me check the options. Option B is 1/10. Hmm. But let me verify again to make sure I didn't make a mistake.Starting over, S_n / n is the average of the first n terms, which is equal to (a1 + a_n)/2. Then S_{2017}/2017 - S_{17}/17 = [(a1 + a_{2017})/2] - [(a1 + a_{17})/2] = [ (a1 + a1 + 2016d)/2 - (a1 + a1 + 16d)/2 ].Calculating that:First term: (2a1 + 2016d)/2 = a1 + 1008dSecond term: (2a1 + 16d)/2 = a1 + 8dSubtracting gives (a1 + 1008d) - (a1 + 8d) = 1000d = 100, so d = 100 / 1000 = 1/10. That's 0.1, which is 1/10. So the answer should be B.But let me check if there's another way to approach this. Maybe using the formula S_n = n/2[2a1 + (n-1)d]. So S_{2017}/2017 would be [2017/2 (2a1 + 2016d)] / 2017 = (2a1 + 2016d)/2, which is the same as before. Similarly for S_17/17. So same result.Alternatively, maybe think of the average as the midpoint of the sequence. For an arithmetic sequence, the average of the first n terms is equal to the average of the first and nth term, which is also equal to the average of all the terms. So, the average when you have 2017 terms would be the middle term, which is the 1009th term (since 2017 is odd). Similarly, the average of 17 terms would be the 9th term. Then the difference between the 1009th term and the 9th term is 100.So, a_1009 - a_9 = 100. But in an arithmetic sequence, a_n = a1 + (n - 1)d. So, a_1009 = a1 + 1008d, and a_9 = a1 + 8d. Therefore, (a1 + 1008d) - (a1 + 8d) = 1000d = 100, so again d = 1/10.So this confirms it again. Therefore, the answer should be B: 1/10.But wait, let me check the options again. The options are A: 1/20, B:1/10, C:10, D:20. So 1/10 is option B. Therefore, the answer is B.But just to be thorough, perhaps there's a trick I'm missing? Let me think. The problem says "the value of d is". So since we derived 1000d = 100, so d = 100/1000 = 1/10. That seems straightforward. The answer is B. So I think that's correct.Final Answerboxed{B}</think>Given an arithmetic sequence ({a_n}) with common difference (d), we need to find (d) given that (frac{S_{2017}}{2017} - frac{S_{17}}{17} = 100).The sum of the first (n) terms of an arithmetic sequence is given by (S_n = frac{n}{2} [2a_1 + (n - 1)d]). The average of the first (n) terms is (frac{S_n}{n} = frac{1}{2} [2a_1 + (n - 1)d]).We calculate the averages for (n = 2017) and (n = 17):[frac{S_{2017}}{2017} = frac{1}{2} [2a_1 + 2016d] = a_1 + 1008d][frac{S_{17}}{17} = frac{1}{2} [2a_1 + 16d] = a_1 + 8d]The difference between these averages is given as 100:[(a_1 + 1008d) - (a_1 + 8d) = 100]Simplifying this equation:[1000d = 100]Solving for (d):[d = frac{100}{1000} = frac{1}{10}]Thus, the value of (d) is (boxed{B}).,<think>Okay, let's tackle this arithmetic sequence problem. So, we have an arithmetic sequence {a_n}, and S_n is the sum of the first n terms. The common difference is d. The problem states that the difference between S_2017 divided by 2017 and S_17 divided by 17 is 100. We need to find the value of d from the given options.First, let me recall the formula for the sum of the first n terms of an arithmetic sequence. The sum S_n is given by:S_n = n/2 [2a_1 + (n - 1)d]Alternatively, it can also be written as:S_n = n(a_1 + a_n)/2Since a_n = a_1 + (n - 1)d, both formulas are equivalent. Maybe using the first one would be more straightforward here because we can express S_n in terms of a_1 and d directly.So, the problem gives us that S_2017 / 2017 - S_17 / 17 = 100. Let me write that down:(S_2017 / 2017) - (S_17 / 17) = 100Let me compute each term separately. Let's start with S_n / n. Using the formula for S_n:S_n / n = [n/2 (2a_1 + (n - 1)d)] / n = (2a_1 + (n - 1)d)/2So, S_n divided by n is equal to (2a_1 + (n - 1)d)/2. That simplifies the problem because now the equation becomes:[ (2a_1 + (2017 - 1)d)/2 ] - [ (2a_1 + (17 - 1)d)/2 ] = 100Let me compute the terms inside the brackets:First term: (2a_1 + 2016d)/2Second term: (2a_1 + 16d)/2Subtracting the second term from the first term:[(2a_1 + 2016d)/2] - [(2a_1 + 16d)/2] = [ (2a_1 + 2016d - 2a_1 - 16d ) ] / 2Simplify the numerator:2a_1 cancels out with -2a_1, so we have (2016d - 16d) / 2 = (2000d)/2 = 1000dSo, the equation becomes:1000d = 100Therefore, solving for d:d = 100 / 1000 = 1/10Hmm, so according to this calculation, d is 1/10, which is option B. Wait, but let me check again to make sure I didn't make any mistakes.Let me go through the steps once more to verify.1. The sum S_n = n/2 [2a_1 + (n - 1)d]2. Therefore, S_n / n = [2a_1 + (n - 1)d]/23. So, S_2017 / 2017 = [2a_1 + 2016d]/24. Similarly, S_17 /17 = [2a_1 + 16d]/25. Subtracting these two: [2a_1 + 2016d]/2 - [2a_1 + 16d]/26. The 2a_1 terms cancel out, leaving (2016d -16d)/2 = 2000d / 2 = 1000d7. So, 1000d = 100 => d = 0.1 or 1/10Yes, that seems correct. Therefore, the answer should be B: 1/10.But wait, let me check if perhaps there was a miscalculation in the difference between the terms. Let's verify:2016d -16d = (2016 -16)d = 2000d. Then divided by 2, so 2000d/2 = 1000d. Then set equal to 100, so 1000d = 100 => d=0.1. Yes, that seems correct.Alternatively, maybe using another approach.Alternatively, recall that the average of the first n terms (which is S_n /n) is equal to the average of the first and nth term. Since in an arithmetic sequence, the average is (a_1 + a_n)/2, which is also equal to the average of all terms.So, S_n /n = (a_1 + a_n)/2.Therefore, S_2017 /2017 = (a_1 + a_2017)/2Similarly, S_17 /17 = (a_1 + a_17)/2So, the difference between these two averages is [(a_1 + a_2017)/2] - [(a_1 + a_17)/2] = [a_2017 - a_17]/2Which is equal to [ (a_1 + 2016d) - (a_1 + 16d) ] /2 = (2000d)/2 = 1000dTherefore, 1000d = 100 => d=0.1Same result. So that's another way to confirm.Alternatively, perhaps think about the difference between the two averages. The average of the first 2017 terms minus the average of the first 17 terms is 100. Since the average in an arithmetic sequence is also the midpoint of the first and last term. So, the difference between the midpoints is 100.But the midpoint of the first 2017 terms is (a_1 + a_2017)/2, and the midpoint of the first 17 terms is (a_1 + a_17)/2. The difference between these midpoints is [(a_1 + a_2017) - (a_1 + a_17)] /2 = (a_2017 - a_17)/2 = ( (a_1 + 2016d) - (a_1 + 16d) ) /2 = (2000d)/2 = 1000d. Therefore, 1000d=100, so d=0.1. So, yes, same result.Therefore, the answer is B: 1/10. So, the common difference is 1/10.But let me check the options again. The options are:A: 1/20B: 1/10C: 10D: 20So, 1/10 is option B. So, the correct answer is B.Alternatively, maybe if I miscalculated, but seems like all approaches lead to 1/10. So, I think that's correct.Final Answerboxed{B}</think>Given an arithmetic sequence ({a_n}) with common difference (d), we know that the sum of the first (n) terms, (S_n), is given by:[ S_n = frac{n}{2} [2a_1 + (n - 1)d] ]We are given that:[ frac{S_{2017}}{2017} - frac{S_{17}}{17} = 100 ]First, we express (frac{S_n}{n}) using the formula for (S_n):[ frac{S_n}{n} = frac{1}{2} [2a_1 + (n - 1)d] ]Applying this to both (S_{2017}/2017) and (S_{17}/17):[ frac{S_{2017}}{2017} = frac{1}{2} [2a_1 + 2016d] ][ frac{S_{17}}{17} = frac{1}{2} [2a_1 + 16d] ]Subtracting these two expressions:[ frac{1}{2} [2a_1 + 2016d] - frac{1}{2} [2a_1 + 16d] = 100 ]Simplifying the left-hand side:[ frac{1}{2} [(2a_1 + 2016d) - (2a_1 + 16d)] = 100 ][ frac{1}{2} [2016d - 16d] = 100 ][ frac{1}{2} cdot 2000d = 100 ][ 1000d = 100 ][ d = frac{100}{1000} = frac{1}{10} ]Thus, the value of (d) is (boxed{B}).]

🔑:From the arithmetic sequence {a_n}, we have: frac{S_n}{n} = a_1 + frac{n-1}{2}d = frac{d}{2}n + (a_1 - frac{1}{2}d), which is an arithmetic sequence.Given frac{S_{2017}}{2017} - frac{S_{17}}{17} = 100,we have frac{d}{2} times 2017 + a_1 - frac{1}{2}d - (frac{d}{2} times 17 + a_1 - frac{1}{2}d) = 100,which simplifies to 10d = 1. Solving this, we find d = frac{1}{10}.Therefore, the correct choice is boxed{B}.This problem examines the properties and sum formula of an arithmetic sequence, testing reasoning and computational skills, and is considered a medium-level question.

🧠:[<think>Okay, so there's this problem about a boat traveling downstream and then needing to return upstream. Let me try to parse through it step by step. Let's see... The river has a main channel in the middle where the water flows at 10 km/h, and near the bank, the water speed is 8 km/h. The boat goes downstream in the middle, covering 360 km in 10 hours. Then it wants to come back along the bank, and we need to figure out how long that will take. Hmm.First, when the boat is going downstream in the middle, its effective speed is the sum of its own speed in still water plus the speed of the current. Conversely, when it goes upstream along the bank, its effective speed will be its own speed minus the current's speed. Wait, but the current speed is different in the middle and near the bank. So the current in the middle is 10 km/h, and near the bank it's 8 km/h. That's important.So let's denote the boat's speed in still water as 'b'. When going downstream in the middle, the total speed would be (b + 10) km/h. They say that in 10 hours, it covers 360 km. So distance equals speed multiplied by time. Therefore, 360 km = (b + 10) km/h * 10 hours. Let me write that equation:360 = 10*(b + 10)Solving for b, we divide both sides by 10:36 = b + 10Subtract 10 from both sides:b = 26 km/hSo the boat's speed in still water is 26 km/h. Now, when it returns along the bank, it's going upstream against a current of 8 km/h. Therefore, its effective speed would be (b - 8) km/h. Plugging in the value of b we found:Effective speed upstream = 26 - 8 = 18 km/hNow, the distance it needs to cover upstream is the same as the distance it went downstream, which is 360 km. Therefore, time taken to go upstream would be distance divided by speed:Time = 360 km / 18 km/h = 20 hoursWait, so the answer is 20 hours? Let me check my steps again to be sure.1. Downstream speed in middle: (b + 10). Time = 10 hours, distance = 360 km. So 10*(b + 10) = 360. Solving gives b = 26. That seems right.2. Upstream along the bank: current is 8 km/h, so effective speed is (26 - 8) = 18 km/h. Time = 360 / 18 = 20. Correct.But wait a second, is the distance really 360 km upstream? The problem says the boat needs to "return to the original place". So yes, the distance should be the same, 360 km. So that seems okay.Wait, but hold on. Rivers can meander, so is the distance upstream the same as downstream? The problem doesn't mention any difference in distance, so I think we can assume the distance is the same. Otherwise, the problem would have specified something. So I think 20 hours is the right answer.But let me think again. Suppose the boat is going downstream with the current, so its speed is increased by the current. Then when going upstream, the current is subtracted. The key here is that the current speed is different depending on where you are in the river. When going downstream, it's in the middle with 10 km/h current, but when going upstream, it's along the bank with 8 km/h current. So the boat's own speed in still water is 26 km/h. Therefore, upstream speed is 26 - 8 = 18 km/h. So time is 360 / 18 = 20. Yeah, that's straightforward.I think that's right. The answer is 20 hours.Final AnswerThe boat will take boxed{20} hours to return to the original place along the bank.</think>The boat travels downstream in the middle of the river where the water flows at a speed of 10 km/h. In 10 hours, it covers 360 km. To find the boat's speed in still water, we use the equation for distance:[360 = 10 times (b + 10)]Solving for ( b ):[36 = b + 10 implies b = 26 text{ km/h}]When returning upstream along the bank where the water flows at 8 km/h, the boat's effective speed is:[26 - 8 = 18 text{ km/h}]The time taken to travel 360 km upstream is:[text{Time} = frac{360}{18} = 20 text{ hours}][boxed{20}],<think>Okay, let's see. The problem is about a boat traveling downstream in a river and then returning upstream along the bank. We need to figure out how long the return trip will take. Hmm, let me read the problem again to make sure I understand all the details.So, the river has a main channel where the water flows at 10 km/h, and near the bank, the water flows at 8 km/h. The boat goes downstream in the middle (main channel) and covers 360 kilometers in 10 hours. Then, it needs to go back to the original place along the bank. We have to find the time it takes for the return trip.Alright, first, when the boat is going downstream in the main channel, its speed relative to the ground (or the banks) is the sum of the boat's speed in still water and the river's speed. Similarly, when going upstream along the bank, its effective speed will be the boat's speed in still water minus the river's speed along the bank.But wait, the problem doesn't mention the boat's speed in still water. Hmm, so maybe we need to find that first using the downstream journey information. Let's denote the boat's speed in still water as 'b' km/h.When going downstream in the main channel, the river's speed is 10 km/h, so the boat's effective speed downstream is (b + 10) km/h. The problem says that in 10 hours, the boat covers 360 km. So, using the formula distance = speed × time, we can write:360 = (b + 10) × 10Let me solve this equation for 'b'. Dividing both sides by 10:36 = b + 10Subtract 10 from both sides:b = 36 - 10 = 26 km/hSo the boat's speed in still water is 26 km/h. Got that.Now, for the return trip, the boat is going upstream along the bank. The river's speed along the bank is given as 8 km/h. Therefore, the boat's effective speed upstream will be its speed in still water minus the river's speed:Effective speed upstream = 26 - 8 = 18 km/hWait a second, hold on. Is that correct? Let me confirm. When going upstream, the boat is moving against the current, so yes, subtract the river's speed from the boat's speed in still water. So 26 - 8 = 18 km/h. That seems right.Now, the distance the boat needs to cover upstream is the same 360 km, right? Because it's returning to the original place. So the distance is 360 km, and the speed is 18 km/h. Therefore, time = distance / speed = 360 / 18.Calculating that: 360 divided by 18. Let's see, 18 times 20 is 360. So 20 hours. Therefore, the return trip will take 20 hours.Wait, that seems straightforward, but let me check again to make sure I didn't make any mistakes.First, downstream speed: boat speed in still water + current. We had 26 + 10 = 36 km/h. Then, 36 km/h times 10 hours = 360 km. That matches the given distance. Good.Then, upstream speed: boat speed in still water minus current along the bank. 26 - 8 = 18 km/h. Distance is 360 km, so time is 360 / 18 = 20 hours. Yes, that seems right.But wait, hold on. The problem mentions that the boat is traveling downstream in the middle of the river where the current is 10 km/h, and then returns along the bank where the current is 8 km/h. So, when returning, is the current 8 km/h or is that something else?Wait, maybe the river's speed along the bank is 8 km/h. So, when the boat is moving upstream along the bank, the current is against it at 8 km/h. Therefore, yes, the effective speed is 26 - 8 = 18 km/h.But let me confirm the problem statement again: "along the bank the water flows at a speed of 8 kilometers per hour." So, the current speed near the bank is 8 km/h, which is slower than the main channel's 10 km/h. That makes sense because near the banks, rivers often have slower currents due to friction.Therefore, when going downstream in the main channel, current is 10 km/h aiding the boat. When returning upstream along the bank, current is 8 km/h opposing the boat.Therefore, the calculation seems correct. Boat speed in still water is 26 km/h. Upstream speed is 18 km/h. Time is 20 hours.Hmm, but let me think again. Is the distance upstream the same as downstream? The problem says "return to the original place," so yes, the distance is the same 360 km. So the answer should be 20 hours.Wait, but hold on. Wait, when the boat goes downstream, it's going with the current of 10 km/h, so the 360 km is relative to the ground. Then, when returning, it's moving against a current of 8 km/h. So, the distance is indeed 360 km, and the effective speed is 18 km/h, so time is 20 hours. That seems correct.Alternatively, maybe the problem is a trick question? But I don't think so. The numbers add up. Let me confirm once more.Boat's downstream speed: 26 + 10 = 36 km/h. 36 km/h * 10 h = 360 km. Correct.Upstream speed: 26 - 8 = 18 km/h. 360 km / 18 km/h = 20 h. Correct.Yes, that seems right. So the answer is 20 hours.Final AnswerThe boat will take boxed{20} hours to return.</think>To determine the time it takes for the boat to return upstream along the bank, we first need to find the boat's speed in still water. When traveling downstream in the main channel, the boat's effective speed is the sum of its speed in still water ((b)) and the river's speed (10 km/h). The boat covers 360 kilometers in 10 hours, so we use the formula for distance:[360 = (b + 10) times 10]Solving for (b):[360 = 10(b + 10) implies 36 = b + 10 implies b = 26 text{ km/h}]Next, for the return trip upstream along the bank, the effective speed is the boat's speed in still water minus the river's speed along the bank (8 km/h):[text{Effective speed upstream} = 26 - 8 = 18 text{ km/h}]The distance to be covered upstream is the same 360 kilometers. Using the formula for time:[text{Time} = frac{text{Distance}}{text{Speed}} = frac{360}{18} = 20 text{ hours}]Thus, the boat will take boxed{20} hours to return.]

🔑:【Solution】Solution: The still water speed of the boat is:begin{array}{l}360 div 10-10, =36-10, =26 text { (km/h); }end{array}To return to the original place, it needs:begin{array}{l}360 div(26-8), =360 div 18, =20 text { (hours); }end{array}Answer: This boat needs 20 hours to return to the original place along the shore. Therefore, the answer is: 20.

🧠:[<think>Okay, so I need to find the smallest positive value of φ such that when the graph of f(x) = sin(2x) + cos(2x) is translated to the right by φ units, the resulting graph is symmetric about the y-axis. Hmm, let's start by recalling some concepts.First, translating a function to the right by φ units means replacing x with (x - φ) in the function. So the translated function would be f(x - φ) = sin[2(x - φ)] + cos[2(x - φ)]. Let me write that out properly:g(x) = sin[2(x - φ)] + cos[2(x - φ)]Now, for the graph of g(x) to be symmetric about the y-axis, it needs to satisfy the condition g(x) = g(-x) for all x. So I need to set up the equation:sin[2(x - φ)] + cos[2(x - φ)] = sin[2(-x - φ)] + cos[2(-x - φ)]Hmm, let's simplify both sides. Starting with the left side:Left Side (LS): sin[2(x - φ)] + cos[2(x - φ)] = sin(2x - 2φ) + cos(2x - 2φ)Right Side (RS): sin[2(-x - φ)] + cos[2(-x - φ)] = sin(-2x - 2φ) + cos(-2x - 2φ)Now, recall that sin(-θ) = -sinθ and cos(-θ) = cosθ. Applying those identities:RS = -sin(2x + 2φ) + cos(2x + 2φ)So the equation becomes:sin(2x - 2φ) + cos(2x - 2φ) = -sin(2x + 2φ) + cos(2x + 2φ)Hmm, this equation must hold for all x, which means the coefficients of the sine and cosine terms on both sides must match. Let me rearrange terms to group similar terms.Bring all terms to the left side:sin(2x - 2φ) + cos(2x - 2φ) + sin(2x + 2φ) - cos(2x + 2φ) = 0Now, let's see if we can use trigonometric identities to combine these terms. Maybe using the sum-to-product formulas?First, let's consider the sine terms: sin(2x - 2φ) + sin(2x + 2φ)Using the identity sin A + sin B = 2 sin[(A+B)/2] cos[(A-B)/2]Here, A = 2x - 2φ, B = 2x + 2φSo:2 sin[( (2x - 2φ) + (2x + 2φ) ) / 2] * cos[( (2x - 2φ) - (2x + 2φ) ) / 2]Simplify:2 sin[ (4x)/2 ] * cos[ (-4φ)/2 ] = 2 sin(2x) * cos(-2φ)But cos(-2φ) = cos(2φ), so this becomes:2 sin(2x) cos(2φ)Now for the cosine terms: cos(2x - 2φ) - cos(2x + 2φ)Using the identity cos A - cos B = -2 sin[(A+B)/2] sin[(A - B)/2]Here, A = 2x - 2φ, B = 2x + 2φSo:-2 sin[ ( (2x - 2φ) + (2x + 2φ) ) / 2 ] * sin[ ( (2x - 2φ) - (2x + 2φ) ) / 2 ]Simplify:-2 sin[ (4x)/2 ] * sin[ (-4φ)/2 ] = -2 sin(2x) * sin(-2φ)But sin(-2φ) = -sin(2φ), so this becomes:-2 sin(2x) * (-sin(2φ)) = 2 sin(2x) sin(2φ)So combining both parts, the entire left side of the equation becomes:2 sin(2x) cos(2φ) + 2 sin(2x) sin(2φ) = 0Factor out 2 sin(2x):2 sin(2x) [ cos(2φ) + sin(2φ) ] = 0Now, this equation must hold for all x. The factor 2 sin(2x) is not identically zero (since sin(2x) isn't zero for all x), so the other factor must be zero:cos(2φ) + sin(2φ) = 0Therefore:cos(2φ) + sin(2φ) = 0Let's solve this equation for φ. Let's divide both sides by cos(2φ) (assuming cos(2φ) ≠ 0):1 + tan(2φ) = 0 ⇒ tan(2φ) = -1Therefore:2φ = arctan(-1) + kπ, where k is any integer.The solutions to tan(θ) = -1 are θ = -π/4 + kπ.So:2φ = -π/4 + kπ ⇒ φ = -π/8 + kπ/2We are looking for the smallest positive φ. Let's find the smallest positive solution by choosing appropriate k.Start with k = 0:φ = -π/8 + 0 = -π/8 (negative, discard)k = 1:φ = -π/8 + π/2 = (-π/8 + 4π/8) = 3π/8k = 2:φ = -π/8 + π = (-π/8 + 8π/8) = 7π/8 (which is larger than 3π/8, so 3π/8 is smaller)Therefore, the smallest positive φ is 3π/8.Wait, let me check if I made an error in calculation.Wait, when k = 1:2φ = -π/4 + π = ( -π/4 + 4π/4 ) = 3π/4 ⇒ φ = 3π/8Similarly, for k = 2:2φ = -π/4 + 2π = ( -π/4 + 8π/4 ) = 7π/4 ⇒ φ = 7π/8Yes, that seems right.Alternatively, maybe I can think of solving tan(2φ) = -1. The solutions for 2φ would be in the second and fourth quadrants where tan is negative.So, 2φ = 3π/4 + kπ ⇒ φ = 3π/8 + kπ/2, where k is an integer.Thus, the smallest positive φ is 3π/8 when k = 0. Wait, but k = 0 gives φ = 3π/8. If k is allowed to be negative, then φ can be negative. But we want the smallest positive, so k = 0 gives the first positive solution. Wait, but let me check.Wait, 2φ = arctan(-1). But arctan(-1) is -π/4, so principal solution is -π/4, but since tan has period π, the general solution is 2φ = -π/4 + kπ. So φ = -π/8 + kπ/2.So when k = 0: φ = -π/8 (negative). k = 1: φ = -π/8 + π/2 = ( -π/8 + 4π/8 ) = 3π/8. So indeed, the smallest positive φ is 3π/8.Wait, but let's verify this answer to make sure.Suppose φ = 3π/8. Let's compute the translated function:g(x) = sin[2(x - 3π/8)] + cos[2(x - 3π/8)]Simplify inside:= sin(2x - 6π/8) + cos(2x - 6π/8) = sin(2x - 3π/4) + cos(2x - 3π/4)Now, check if g(x) is even, i.e., g(-x) = g(x).Compute g(-x):sin[2(-x) - 3π/4] + cos[2(-x) - 3π/4] = sin(-2x - 3π/4) + cos(-2x - 3π/4) = -sin(2x + 3π/4) + cos(2x + 3π/4)We need this to equal g(x) = sin(2x - 3π/4) + cos(2x - 3π/4)So, equate:-sin(2x + 3π/4) + cos(2x + 3π/4) = sin(2x - 3π/4) + cos(2x - 3π/4)Let me use angle addition formulas.First, compute sin(2x - 3π/4) and cos(2x - 3π/4):sin(2x - 3π/4) = sin2x cos(3π/4) - cos2x sin(3π/4) = sin2x (-√2/2) - cos2x (√2/2) = -√2/2 sin2x - √2/2 cos2xcos(2x - 3π/4) = cos2x cos(3π/4) + sin2x sin(3π/4) = cos2x (-√2/2) + sin2x (√2/2) = -√2/2 cos2x + √2/2 sin2xSo adding them together:sin(2x - 3π/4) + cos(2x - 3π/4) = [ -√2/2 sin2x - √2/2 cos2x ] + [ -√2/2 cos2x + √2/2 sin2x ] = (-√2/2 sin2x + √2/2 sin2x) + (-√2/2 cos2x - √2/2 cos2x ) = 0 - √2 cos2x = -√2 cos2xNow compute the left side: -sin(2x + 3π/4) + cos(2x + 3π/4)First, sin(2x + 3π/4) = sin2x cos(3π/4) + cos2x sin(3π/4) = sin2x (-√2/2) + cos2x (√2/2) = -√2/2 sin2x + √2/2 cos2xSimilarly, cos(2x + 3π/4) = cos2x cos(3π/4) - sin2x sin(3π/4) = cos2x (-√2/2) - sin2x (√2/2) = -√2/2 cos2x - √2/2 sin2xSo -sin(2x + 3π/4) + cos(2x + 3π/4) = -[ -√2/2 sin2x + √2/2 cos2x ] + [ -√2/2 cos2x - √2/2 sin2x ]Simplify:= √2/2 sin2x - √2/2 cos2x - √2/2 cos2x - √2/2 sin2xCombine like terms:(√2/2 sin2x - √2/2 sin2x) + (-√2/2 cos2x - √2/2 cos2x ) = 0 - √2 cos2x = -√2 cos2xSo both sides equal -√2 cos2x. Therefore, g(-x) = g(x) for all x, which means the translated function is indeed symmetric about the y-axis. Therefore, φ = 3π/8 is correct.So the smallest positive value of φ is 3π/8.Final AnswerThe smallest positive value of (varphi) is (boxed{dfrac{3pi}{8}}).</think>To find the smallest positive value of (varphi) such that the graph of (f(x) = sin(2x) + cos(2x)) translated to the right by (varphi) units is symmetric about the (y)-axis, we start by considering the translated function (g(x) = sin[2(x - varphi)] + cos[2(x - varphi)]). For (g(x)) to be symmetric about the (y)-axis, it must satisfy (g(x) = g(-x)) for all (x).Substituting the translated function into the symmetry condition, we get:[sin(2(x - varphi)) + cos(2(x - varphi)) = sin(2(-x - varphi)) + cos(2(-x - varphi))]Simplifying both sides using trigonometric identities, we have:[sin(2x - 2varphi) + cos(2x - 2varphi) = -sin(2x + 2varphi) + cos(2x + 2varphi)]Rearranging and combining terms using sum-to-product identities, we obtain:[2 sin(2x) cos(2varphi) + 2 sin(2x) sin(2varphi) = 0]Factoring out common terms, we get:[2 sin(2x) [cos(2varphi) + sin(2varphi)] = 0]Since (sin(2x)) is not identically zero, we must have:[cos(2varphi) + sin(2varphi) = 0]Solving this equation, we divide by (cos(2varphi)) (assuming (cos(2varphi) neq 0)):[1 + tan(2varphi) = 0 implies tan(2varphi) = -1]The general solution for (tan(2varphi) = -1) is:[2varphi = -frac{pi}{4} + kpi implies varphi = -frac{pi}{8} + frac{kpi}{2} quad text{for integer } k]To find the smallest positive (varphi), we test integer values of (k):- For (k = 1), (varphi = -frac{pi}{8} + frac{pi}{2} = frac{3pi}{8})Thus, the smallest positive value of (varphi) is (boxed{dfrac{3pi}{8}}).,<think>Okay, let's see. I need to find the smallest positive value of φ such that when the graph of f(x) = sin(2x) + cos(2x) is translated to the right by φ units, the resulting graph is symmetric about the y-axis. Hmm, symmetry about the y-axis means that for the translated function, say g(x), it must satisfy g(x) = g(-x) for all x. First, let's recall that translating a function to the right by φ units replaces x with (x - φ) in the function. So the translated function would be:g(x) = f(x - φ) = sin[2(x - φ)] + cos[2(x - φ)]Simplify that:sin(2x - 2φ) + cos(2x - 2φ)Now, we need this function g(x) to be even, meaning g(x) = g(-x). So let's set up that equation:sin(2x - 2φ) + cos(2x - 2φ) = sin(-2x - 2φ) + cos(-2x - 2φ)Hmm, let's simplify the right side. Remember that sin(-θ) = -sinθ and cos(-θ) = cosθ. So:sin(-2x - 2φ) = -sin(2x + 2φ)cos(-2x - 2φ) = cos(2x + 2φ)Therefore, the right side becomes:-sin(2x + 2φ) + cos(2x + 2φ)So the equation is:sin(2x - 2φ) + cos(2x - 2φ) = -sin(2x + 2φ) + cos(2x + 2φ)This equation must hold for all x. That seems complicated. Maybe there's a trigonometric identity that can help simplify both sides.Alternatively, maybe it's easier to first express the original function f(x) = sin(2x) + cos(2x) in a different form, like a single sine or cosine function. Since combining sine and cosine terms can be done using amplitude-phase form. Let's try that.Recall that A sinθ + B cosθ = C sin(θ + φ) or C cos(θ + φ'), depending on the phase shift. Let's use the identity:A sinθ + B cosθ = √(A² + B²) sin(θ + α), where α = arctan(B/A) or something like that. Wait, let me check.Actually, the formula is:A sinθ + B cosθ = C sin(θ + φ), where C = √(A² + B²) and φ = arctan(B/A) if we consider the right phase shift. Wait, maybe I should use cosine:Alternatively, it can be written as C cos(θ - δ), where δ is another angle. Let me recall.The identity is:A sinθ + B cosθ = C cos(θ - δ), where C = √(A² + B²) and tanδ = A/B.Wait, let me verify. Let's expand C cos(θ - δ):C cosθ cosδ + C sinθ sinδ. Comparing with A sinθ + B cosθ, we have:B = C cosδ and A = C sinδ. Therefore, tanδ = A/B, so δ = arctan(A/B). Then C = √(A² + B²).So in this case, A = 1 (coefficient of sin2x) and B = 1 (coefficient of cos2x). Therefore:C = √(1 + 1) = √2, and δ = arctan(1/1) = π/4.Therefore, f(x) = sin(2x) + cos(2x) can be written as √2 cos(2x - π/4). Let me check that:√2 cos(2x - π/4) = √2 [cos2x cos(π/4) + sin2x sin(π/4)] = √2 [ (cos2x)(√2/2) + (sin2x)(√2/2) ] = √2*(√2/2)(sin2x + cos2x) = (2/2)(sin2x + cos2x) = sin2x + cos2x. Perfect, that works.So f(x) = √2 cos(2x - π/4). Therefore, translating this function to the right by φ units gives:g(x) = √2 cos[2(x - φ) - π/4] = √2 cos(2x - 2φ - π/4)We need this function to be even, so g(x) = g(-x). So:√2 cos(2x - 2φ - π/4) = √2 cos(-2x - 2φ - π/4)Since cosine is even, cos(-θ) = cosθ, so the right side becomes:√2 cos(2x + 2φ + π/4)Therefore, the equation is:cos(2x - 2φ - π/4) = cos(2x + 2φ + π/4)For this equality to hold for all x, the arguments of the cosine functions must differ by a multiple of 2π or be negatives plus a multiple of 2π. Remember that cosα = cosβ if and only if α = β + 2πk or α = -β + 2πk for some integer k.Therefore, for all x, we must have:2x - 2φ - π/4 = 2x + 2φ + π/4 + 2πkor2x - 2φ - π/4 = - (2x + 2φ + π/4) + 2πkLet's consider the first case:2x - 2φ - π/4 = 2x + 2φ + π/4 + 2πkSubtract 2x from both sides:-2φ - π/4 = 2φ + π/4 + 2πkBring terms with φ to one side:-2φ - 2φ = π/4 + π/4 + 2πk-4φ = π/2 + 2πkThus,φ = - (π/2 + 2πk)/4 = -π/8 - πk/2Since we need the smallest positive φ, let's see for integers k:If k = -1, φ = -π/8 - (-π/2) = -π/8 + π/2 = 3π/8 ≈ 1.178, which is positive.If k = 0, φ = -π/8, which is negative.For k = -2, φ = -π/8 - (-π) = -π/8 + π = 7π/8 ≈ 2.748, which is larger than 3π/8.So the smallest positive φ from this case is 3π/8. But let's check the other case to see if we get a smaller positive φ.Second case:2x - 2φ - π/4 = -2x - 2φ - π/4 + 2πkBring all terms to one side:2x - 2φ - π/4 + 2x + 2φ + π/4 = 2πkSimplify:4x = 2πkTherefore:x = (πk)/2But this must hold for all x, which is only possible if k = 0, leading to x = 0, but that's only true for x = 0, not for all x. Therefore, this case doesn't provide a solution valid for all x. Therefore, the only valid solutions come from the first case.Thus, the smallest positive φ is 3π/8.Wait, but let me verify this result. Let's check if shifting the original function by 3π/8 to the right results in an even function.Original function f(x) = sin2x + cos2x. Shifted right by 3π/8:g(x) = sin[2(x - 3π/8)] + cos[2(x - 3π/8)] = sin(2x - 3π/4) + cos(2x - 3π/4)Let's compute g(-x):sin(-2x - 3π/4) + cos(-2x - 3π/4) = -sin(2x + 3π/4) + cos(2x + 3π/4)We need to check if this equals g(x) = sin(2x - 3π/4) + cos(2x - 3π/4)So let's compute:sin(2x - 3π/4) + cos(2x - 3π/4) vs. -sin(2x + 3π/4) + cos(2x + 3π/4)Using trigonometric identities:First, sin(2x - 3π/4) = sin2x cos(3π/4) - cos2x sin(3π/4) = sin2x*(-√2/2) - cos2x*(√2/2) = (-√2/2)(sin2x + cos2x)Similarly, cos(2x - 3π/4) = cos2x cos(3π/4) + sin2x sin(3π/4) = cos2x*(-√2/2) + sin2x*(√2/2) = (√2/2)(sin2x - cos2x)Therefore, g(x) = (-√2/2)(sin2x + cos2x) + (√2/2)(sin2x - cos2x) = (-√2/2 sin2x - √2/2 cos2x) + (√2/2 sin2x - √2/2 cos2x) = (-√2/2 + √2/2) sin2x + (-√2/2 - √2/2) cos2x = 0*sin2x - √2 cos2x = -√2 cos2xWait, that's interesting. So shifting f(x) by 3π/8 to the right gives g(x) = -√2 cos2x. Let's check if that's even.g(-x) = -√2 cos(-2x) = -√2 cos2x = g(x). Yes, because cosine is even. So indeed, g(-x) = g(x). Therefore, shifting by 3π/8 makes the function even. Therefore, 3π/8 is the correct answer.But wait, the problem says "the smallest positive value of φ". Since 3π/8 is approximately 1.178, which is positive. Let me check if there's a smaller positive φ. Suppose k = -1 gives 3π/8, k = -2 gives 7π/8, which is larger, so 3π/8 is indeed the smallest positive. So the answer should be 3π/8.But let's verify using another approach. Maybe we can use the original function f(x) = sin2x + cos2x. If we shift it right by φ, then for the shifted function to be even, f(φ - x) = f(φ + x) for all x. Wait, actually, if you shift right by φ, then the function is f(x - φ). For it to be even, we need f(-x - φ) = f(x - φ) for all x. Wait, no, maybe better to think of the shifted function as g(x) = f(x - φ). For g(x) to be even, g(x) = g(-x) => f(x - φ) = f(-x - φ) for all x.So f(x - φ) = f(-x - φ) for all x. Let's substitute y = x - φ. Then x = y + φ. So substituting into the equation:f(y) = f(- (y + φ) - φ) = f(-y - 2φ)Therefore, we have f(y) = f(-y - 2φ) for all y. So this must hold for all y. Therefore, the function f is symmetric about the point y = -φ. Wait, but f is periodic, so maybe this condition imposes a specific relationship.Given f(y) = sin2y + cos2y, and f(-y - 2φ) = sin[2(-y - 2φ)] + cos[2(-y - 2φ)] = sin(-2y - 4φ) + cos(-2y - 4φ) = -sin(2y + 4φ) + cos(2y + 4φ)Therefore, the equation f(y) = f(-y - 2φ) becomes:sin2y + cos2y = -sin(2y + 4φ) + cos(2y + 4φ)This must hold for all y. Let's rearrange:sin2y + cos2y + sin(2y + 4φ) - cos(2y + 4φ) = 0Hmm, this seems similar to the previous equation. Maybe using the same approach as before, express both sides in terms of combined sine and cosine.Alternatively, express f(y) as √2 cos(2y - π/4) as before. Then f(-y - 2φ) = √2 cos[2(-y - 2φ) - π/4] = √2 cos(-2y - 4φ - π/4) = √2 cos(2y + 4φ + π/4), since cosine is even.Therefore, the equation is:√2 cos(2y - π/4) = √2 cos(2y + 4φ + π/4)Divide both sides by √2:cos(2y - π/4) = cos(2y + 4φ + π/4)Again, using the identity that cosA = cosB implies A = B + 2πk or A = -B + 2πk.So either:2y - π/4 = 2y + 4φ + π/4 + 2πkor2y - π/4 = - (2y + 4φ + π/4) + 2πkFirst case:2y - π/4 = 2y + 4φ + π/4 + 2πkSubtract 2y from both sides:-π/4 = 4φ + π/4 + 2πkThen:-π/4 - π/4 - 2πk = 4φ-π/2 - 2πk = 4φThus,φ = -π/8 - πk/2Same as before. So φ = -π/8 - πk/2. For positive φ, set k = -1: φ = -π/8 + π/2 = 3π/8; k = -2: φ = -π/8 + π = 7π/8, etc. So same result.Second case:2y - π/4 = -2y - 4φ - π/4 + 2πkBring terms together:2y + 2y = -4φ - π/4 + π/4 + 2πk4y = -4φ + 2πkThus,y = -φ + πk/2But this must hold for all y, which is only possible if the coefficients of y are zero and the constants also. But here, we have 4y = -4φ + 2πk. The left side has a term with y, and the right side doesn't, unless the coefficient of y is zero, but 4 ≠ 0. Therefore, no solution in this case. Hence, only solutions from the first case.Therefore, the smallest positive φ is 3π/8. Let me check once again with another perspective. Suppose we have the original function f(x) = sin2x + cos2x, which we can write as √2 cos(2x - π/4). Shifting it right by φ gives √2 cos(2(x - φ) - π/4) = √2 cos(2x - 2φ - π/4). For this to be even, it must satisfy √2 cos(2x - 2φ - π/4) = √2 cos(-2x - 2φ - π/4). Since cosine is even, this simplifies to √2 cos(2x - 2φ - π/4) = √2 cos(2x + 2φ + π/4). Therefore, the arguments must satisfy either 2x - 2φ - π/4 = 2x + 2φ + π/4 + 2πk or 2x - 2φ - π/4 = - (2x + 2φ + π/4) + 2πk. As before, the first case leads to φ = -π/8 - πk/2 and the second case leads to an inconsistency unless x is specific. Hence, φ = 3π/8 when k = -1.Yes, so all approaches confirm that the smallest positive φ is 3π/8. Therefore, the answer is 3π/8.Final AnswerThe smallest positive value of (varphi) is (boxed{dfrac{3pi}{8}}).</think>To find the smallest positive value of (varphi) such that translating the graph of (f(x) = sin 2x + cos 2x) to the right by (varphi) units results in a graph symmetric about the (y)-axis, we start by expressing the function in a more convenient form. The function (f(x) = sin 2x + cos 2x) can be rewritten using the amplitude-phase form:[f(x) = sqrt{2} cosleft(2x - frac{pi}{4}right)]When this function is translated to the right by (varphi) units, the resulting function is:[g(x) = sqrt{2} cosleft(2(x - varphi) - frac{pi}{4}right) = sqrt{2} cosleft(2x - 2varphi - frac{pi}{4}right)]For (g(x)) to be symmetric about the (y)-axis, it must satisfy (g(x) = g(-x)). This implies:[sqrt{2} cosleft(2x - 2varphi - frac{pi}{4}right) = sqrt{2} cosleft(-2x - 2varphi - frac{pi}{4}right)]Using the even property of the cosine function, (cos(-theta) = cos(theta)), we get:[cosleft(2x - 2varphi - frac{pi}{4}right) = cosleft(2x + 2varphi + frac{pi}{4}right)]For this equality to hold for all (x), the arguments of the cosine functions must differ by a multiple of (2pi) or be negatives plus a multiple of (2pi). This gives us two cases:1. (2x - 2varphi - frac{pi}{4} = 2x + 2varphi + frac{pi}{4} + 2kpi)2. (2x - 2varphi - frac{pi}{4} = -2x - 2varphi - frac{pi}{4} + 2kpi)Solving the first case:[-2varphi - frac{pi}{4} = 2varphi + frac{pi}{4} + 2kpi][-4varphi = frac{pi}{2} + 2kpi][varphi = -frac{pi}{8} - frac{kpi}{2}]For the smallest positive (varphi), we choose (k = -1):[varphi = -frac{pi}{8} + frac{pi}{2} = frac{3pi}{8}]The second case does not provide a valid solution as it leads to a contradiction unless specific values of (x) are considered, which is not valid for all (x).Thus, the smallest positive value of (varphi) is (boxed{dfrac{3pi}{8}}).]

🔑:Firstly, applying the sum-to-product identity, we can rewrite the original function as: f(x) = sin 2x + cos 2x = sqrt{2} sinleft(2x + frac{pi}{4}right) To find the new function after the graph is translated to the right by varphi units, we replace x with (x-varphi): y = sqrt{2} sinleft(2(x-varphi) + frac{pi}{4}right) = sqrt{2} sinleft(2x - 2varphi + frac{pi}{4}right) The resulting graph is symmetric about the y-axis. This means that the function representing the graph should be even, and sin(2x - 2varphi + frac{pi}{4}) should be equal to sin(2(-x) - 2varphi + frac{pi}{4}), which is sin(-2x - 2varphi + frac{pi}{4}). For the sine function, which is odd, this implies: sin(2x - 2varphi + frac{pi}{4}) = -sin(2x + 2varphi - frac{pi}{4}) For this equation to hold for all x, we must have: - 2varphi + frac{pi}{4} = kpi - frac{pi}{4} where k is an integer, because sin(alpha) = -sin(alpha + kpi) for integer k. Solving for varphi, we get: -2varphi = kpi - frac{pi}{2} 2varphi = -kpi + frac{pi}{2} varphi = frac{-kpi + frac{pi}{2}}{2} To find the smallest positive varphi, we can set k to -1 (since k = 0 will give us varphi = frac{pi}{4}, which is not the minimum positive solution we are looking for): varphi = frac{-(-1)pi + frac{pi}{2}}{2} = frac{pi + frac{pi}{2}}{2} = frac{3pi}{4} div 2 = frac{3pi}{8} Therefore, the smallest positive value of varphi is boxed{frac{3pi}{8}}.

🧠:[<think>Alright, let's tackle this system of equations. Hmm, the first equation is ( x + y + frac{9}{x} + frac{4}{y} = 10 ), and the second one is ( (x^2 + 9)(y^2 + 4) = 24xy ). Okay, both equations involve ( x ) and ( y ), and there are fractions and products. Maybe substitution or some clever manipulation will help here.Looking at the first equation: ( x + frac{9}{x} + y + frac{4}{y} = 10 ). It seems like each variable is paired with a term involving its reciprocal. Maybe I can let ( a = x + frac{9}{x} ) and ( b = y + frac{4}{y} ), so the first equation becomes ( a + b = 10 ). But I'm not sure if that helps yet. Let's see what the second equation looks like.The second equation is ( (x^2 + 9)(y^2 + 4) = 24xy ). Let's expand the left side: ( x^2y^2 + 4x^2 + 9y^2 + 36 = 24xy ). Hmm, not sure if expanding is the way to go. Maybe there's a way to relate this to the terms in the first equation.Wait, the first equation has ( x + 9/x ) and the second equation has ( x^2 + 9 ). Notice that ( x^2 + 9 = x left( x + frac{9}{x} right) ). Similarly, ( y^2 + 4 = y left( y + frac{4}{y} right) ). So maybe substituting those into the second equation could be useful.Let me try that. Let ( a = x + frac{9}{x} ) and ( b = y + frac{4}{y} ), then ( x^2 + 9 = x cdot a ) and ( y^2 + 4 = y cdot b ). So the second equation becomes ( (x cdot a)(y cdot b) = 24xy ), which simplifies to ( xyab = 24xy ). Assuming ( x ) and ( y ) are not zero (since they are denominators in the original equations), we can divide both sides by ( xy ) to get ( ab = 24 ).Now, from the first equation, we have ( a + b = 10 ). So now we have a system in terms of ( a ) and ( b ):[begin{cases}a + b = 10 ab = 24end{cases}]This is a simpler system. Solving this, we can recognize that ( a ) and ( b ) are the roots of the quadratic equation ( t^2 - 10t + 24 = 0 ). Factoring this, we get ( (t - 6)(t - 4) = 0 ), so ( t = 6 ) or ( t = 4 ). Therefore, ( a ) and ( b ) are 6 and 4 in some order.So, there are two possibilities:1. ( a = 6 ) and ( b = 4 )2. ( a = 4 ) and ( b = 6 )Let's consider each case.Case 1: ( a = 6 ) and ( b = 4 )- For ( a = 6 ): ( x + frac{9}{x} = 6 ) Multiply both sides by ( x ): ( x^2 + 9 = 6x ) Rearranged: ( x^2 - 6x + 9 = 0 ) This factors as ( (x - 3)^2 = 0 ), so ( x = 3 )- For ( b = 4 ): ( y + frac{4}{y} = 4 ) Multiply both sides by ( y ): ( y^2 + 4 = 4y ) Rearranged: ( y^2 - 4y + 4 = 0 ) This factors as ( (y - 2)^2 = 0 ), so ( y = 2 )So one solution is ( x = 3 ), ( y = 2 ).Case 2: ( a = 4 ) and ( b = 6 )- For ( a = 4 ): ( x + frac{9}{x} = 4 ) Multiply both sides by ( x ): ( x^2 + 9 = 4x ) Rearranged: ( x^2 - 4x + 9 = 0 ) The discriminant here is ( (-4)^2 - 4 cdot 1 cdot 9 = 16 - 36 = -20 ), which is negative. So no real solutions for ( x ) in this case.- For ( b = 6 ): ( y + frac{4}{y} = 6 ) Multiply both sides by ( y ): ( y^2 + 4 = 6y ) Rearranged: ( y^2 - 6y + 4 = 0 ) The discriminant is ( (-6)^2 - 4 cdot 1 cdot 4 = 36 - 16 = 20 ), which is positive. Solutions are ( y = frac{6 pm sqrt{20}}{2} = 3 pm sqrt{5} ). However, since ( a = 4 ) gave no real solutions, this case doesn't contribute any real solutions.Therefore, the only real solution is ( x = 3 ), ( y = 2 ).Wait, but I should check if this solution actually satisfies both original equations. Let me verify.First equation: ( 3 + 2 + frac{9}{3} + frac{4}{2} = 3 + 2 + 3 + 2 = 10 ). Yes, that's correct.Second equation: ( (3^2 + 9)(2^2 + 4) = (9 + 9)(4 + 4) = (18)(8) = 144 ). On the right side: ( 24 cdot 3 cdot 2 = 24 cdot 6 = 144 ). So that's correct too.Therefore, ( x = 3 ), ( y = 2 ) is indeed a solution. Since the other case didn't yield real solutions, this is the only solution.Wait a second, but systems of equations can sometimes have multiple solutions. Did I miss anything here? Let me think.In Case 1, both equations for ( x ) and ( y ) gave unique solutions. In Case 2, the equations didn't have real solutions. So the only real solution is ( (3, 2) ). Let me check if there could be other solutions.Alternatively, maybe I should check if there are other possibilities when setting ( a ) and ( b ). But since the quadratic only gives two roots, 6 and 4, and we considered both permutations, I think that's all. So I think the solution is unique.Alternatively, maybe there's another approach. Let me try substituting ( x = 3 ) and ( y = 2 ) into the original equations to make sure.First equation: ( 3 + 2 + 9/3 + 4/2 = 3 + 2 + 3 + 2 = 10 ). Correct.Second equation: ( (9 + 9)(4 + 4) = 18 * 8 = 144 ), and ( 24 * 3 * 2 = 144 ). Correct.So, seems solid. Maybe that's the only solution.Alternatively, let's consider if ( x ) and ( y ) can be negative. Wait, in the original equations, denominators are ( x ) and ( y ), so ( x ) and ( y ) can't be zero, but they can be negative. However, the problem didn't specify if we're looking for real solutions only, but since the first case gave us positive integers, maybe the solutions are positive. Let's check if negative solutions could exist.Suppose ( x ) is negative, then ( x + 9/x ). For example, if ( x = -3 ), then ( -3 + 9/(-3) = -3 -3 = -6 ). Similarly, if ( x = -a ), then ( x + 9/x = -a -9/a ). Similarly for ( y ). But then, in the first equation, if ( x ) is negative, ( x + 9/x ) would be negative, and same for ( y ). Adding them up might not reach 10. Let's see.Suppose ( x ) and ( y ) are negative. Let ( x = -u ), ( y = -v ), where ( u, v > 0 ). Then the first equation becomes:( -u - v + frac{9}{-u} + frac{4}{-v} = -u - v - frac{9}{u} - frac{4}{v} = 10 ). So ( - (u + v + 9/u + 4/v) = 10 ), which would mean ( u + v + 9/u + 4/v = -10 ), but since ( u, v > 0 ), the left side is positive, which can't equal -10. So negative values of ( x ) or ( y ) would lead to contradictions in the first equation. Therefore, ( x ) and ( y ) must be positive. So we don't need to consider negative solutions.Therefore, the only real solution is ( x = 3 ), ( y = 2 ).Final AnswerThe solution to the system is boxed{(3, 2)}.</think>Given the system of equations:[left{begin{array}{l}x + y + frac{9}{x} + frac{4}{y} = 10 (x^{2} + 9)(y^{2} + 4) = 24xyend{array}right.]First, we introduce substitutions ( a = x + frac{9}{x} ) and ( b = y + frac{4}{y} ). This transforms the first equation into:[a + b = 10]For the second equation, we note that ( x^2 + 9 = x left( x + frac{9}{x} right) = x cdot a ) and ( y^2 + 4 = y left( y + frac{4}{y} right) = y cdot b ). Substituting these into the second equation gives:[(x cdot a)(y cdot b) = 24xy]Simplifying, we get:[ab = 24]We now solve the system:[begin{cases}a + b = 10 ab = 24end{cases}]The roots of the quadratic equation ( t^2 - 10t + 24 = 0 ) are ( t = 6 ) and ( t = 4 ). Thus, ( (a, b) ) can be ( (6, 4) ) or ( (4, 6) ).Case 1: ( a = 6 ) and ( b = 4 )- For ( x + frac{9}{x} = 6 ): [ x^2 - 6x + 9 = 0 implies (x - 3)^2 = 0 implies x = 3 ]- For ( y + frac{4}{y} = 4 ): [ y^2 - 4y + 4 = 0 implies (y - 2)^2 = 0 implies y = 2 ]Case 2: ( a = 4 ) and ( b = 6 )- For ( x + frac{9}{x} = 4 ): [ x^2 - 4x + 9 = 0 ] The discriminant is negative, so no real solutions.- For ( y + frac{4}{y} = 6 ): [ y^2 - 6y + 4 = 0 ] The discriminant is positive, but since ( x ) has no real solutions, this case is discarded.Verifying the solution ( (x, y) = (3, 2) ):- First equation: ( 3 + 2 + frac{9}{3} + frac{4}{2} = 10 )- Second equation: ( (3^2 + 9)(2^2 + 4) = 144 ) and ( 24 cdot 3 cdot 2 = 144 )Thus, the solution to the system is boxed{(3, 2)}.,<think>Okay, so I need to solve this system of equations:First equation: ( x + y + frac{9}{x} + frac{4}{y} = 10 )Second equation: ( (x^2 + 9)(y^2 + 4) = 24xy )Hmm, let me see. Both equations have ( x ) and ( y ) in them, and also their reciprocals in the first equation. The second equation is a product of two quadratics equal to 24xy. This looks a bit complicated, but maybe there's a substitution or some manipulation that can simplify things.Starting with the first equation: ( x + y + frac{9}{x} + frac{4}{y} = 10 ). Let's see if I can group terms here. The terms ( x + frac{9}{x} ) and ( y + frac{4}{y} ) seem similar. Maybe I can set ( a = x + frac{9}{x} ) and ( b = y + frac{4}{y} ), so the first equation becomes ( a + b = 10 ). That might be helpful later.Now, looking at the second equation: ( (x^2 + 9)(y^2 + 4) = 24xy ). Let me expand the left side first.Expanding ( (x^2 + 9)(y^2 + 4) ):( x^2 y^2 + 4x^2 + 9y^2 + 36 = 24xy )Hmm, so ( x^2 y^2 + 4x^2 + 9y^2 + 36 = 24xy ). That seems messy. Maybe there's a way to relate this to the first equation? Let's see.In the first equation, we have terms like ( x + frac{9}{x} ). If I multiply ( x + frac{9}{x} ) by ( x ), I get ( x^2 + 9 ). Similarly, multiplying ( y + frac{4}{y} ) by ( y ) gives ( y^2 + 4 ). Interesting! So, ( x^2 + 9 = x left( x + frac{9}{x} right ) = x cdot a ), and ( y^2 + 4 = y left( y + frac{4}{y} right ) = y cdot b ). So, substituting these into the second equation:( (x cdot a)(y cdot b) = 24xy )Simplifying left side: ( xyab = 24xy )Assuming ( xy neq 0 ), we can divide both sides by ( xy ):( ab = 24 )So now, from the first equation we had ( a + b = 10 ), and now we have ( ab = 24 ). So, this reduces to a system of equations in ( a ) and ( b ):( a + b = 10 )( ab = 24 )This is a quadratic equation in terms of either ( a ) or ( b ). Let me solve for ( a ) and ( b ).Let ( a ) and ( b ) be roots of the equation ( t^2 - 10t + 24 = 0 ).Factoring: ( (t - 6)(t - 4) = 0 ), so ( t = 6 ) or ( t = 4 ).So, the possible pairs ( (a, b) ) are ( (6, 4) ) and ( (4, 6) ).Therefore, either:Case 1: ( a = 6 ) and ( b = 4 )OrCase 2: ( a = 4 ) and ( b = 6 )Now, recalling that ( a = x + frac{9}{x} ) and ( b = y + frac{4}{y} ), we can set up equations for each case.Case 1: ( a = 6 ), ( b = 4 )So,1. ( x + frac{9}{x} = 6 )2. ( y + frac{4}{y} = 4 )Let's solve each equation.Starting with equation 1: Multiply both sides by ( x ):( x^2 + 9 = 6x )( x^2 - 6x + 9 = 0 )This factors as ( (x - 3)^2 = 0 ), so ( x = 3 ).Equation 2: Multiply both sides by ( y ):( y^2 + 4 = 4y )( y^2 - 4y + 4 = 0 )This factors as ( (y - 2)^2 = 0 ), so ( y = 2 ).Therefore, one solution is ( x = 3 ), ( y = 2 ).Case 2: ( a = 4 ), ( b = 6 )So,1. ( x + frac{9}{x} = 4 )2. ( y + frac{4}{y} = 6 )Again, solving each equation.Equation 1: Multiply by ( x ):( x^2 + 9 = 4x )( x^2 - 4x + 9 = 0 )Discriminant: ( (-4)^2 - 4(1)(9) = 16 - 36 = -20 ), which is negative. So, no real solutions here.Equation 2: Multiply by ( y ):( y^2 + 4 = 6y )( y^2 - 6y + 4 = 0 )Discriminant: ( 36 - 16 = 20 ), so solutions are ( y = frac{6 pm sqrt{20}}{2} = frac{6 pm 2sqrt{5}}{2} = 3 pm sqrt{5} ).However, since equation 1 in this case has no real solutions, the entire case 2 doesn't yield real solutions for ( x ) and ( y ). Therefore, case 2 is invalid.Hence, the only real solution is ( x = 3 ), ( y = 2 ).But wait, let me check if this solution satisfies both original equations.First equation: ( 3 + 2 + frac{9}{3} + frac{4}{2} = 3 + 2 + 3 + 2 = 10 ). Correct.Second equation: ( (3^2 + 9)(2^2 + 4) = (9 + 9)(4 + 4) = 18 * 8 = 144 ). And 24xy = 24*3*2 = 144. So that's correct.Therefore, the solution is ( x = 3 ), ( y = 2 ).Wait, but the system is symmetric in some terms. Could there be other solutions where x and y are negative? Let me check if there are other possibilities.But let's see, the original equations have terms like ( frac{9}{x} ) and ( frac{4}{y} ). If x or y is negative, those terms would be negative, but adding them to x and y. Let me see if negative solutions are possible.Suppose in case 1, x is negative. Let's suppose x is negative, but y is positive? Wait, but in case 1, we had x + 9/x = 6. If x is negative, then x + 9/x would be negative - 9/|x|, which is negative, so it can't be equal to 6. Similarly, y + 4/y = 4. If y is negative, same thing. So, x and y must be positive.Hence, x and y must be positive real numbers. Therefore, negative solutions are not possible.Therefore, the only solution is x=3, y=2.But just to make sure, let's check case 2 again. Even though equation 1 had no real solutions, maybe if we consider complex numbers, but the problem probably wants real solutions. So, yes, only case 1 works.Therefore, the solution is (3, 2).Final AnswerThe solution to the system is boxed{(3, 2)}.</think>Given the system of equations:[left{begin{array}{l}x + y + frac{9}{x} + frac{4}{y} = 10 (x^2 + 9)(y^2 + 4) = 24xyend{array}right.]We start by introducing substitutions ( a = x + frac{9}{x} ) and ( b = y + frac{4}{y} ). This transforms the first equation into:[a + b = 10]For the second equation, we note that ( x^2 + 9 = x left( x + frac{9}{x} right) = x cdot a ) and ( y^2 + 4 = y left( y + frac{4}{y} right) = y cdot b ). Substituting these into the second equation gives:[(x cdot a)(y cdot b) = 24xy]Dividing both sides by ( xy ) (assuming ( xy neq 0 )) results in:[ab = 24]We now solve the system ( a + b = 10 ) and ( ab = 24 ). The quadratic equation ( t^2 - 10t + 24 = 0 ) factors as ( (t - 6)(t - 4) = 0 ), giving roots ( t = 6 ) and ( t = 4 ). Thus, the possible pairs ((a, b)) are ((6, 4)) and ((4, 6)).Case 1: ( a = 6 ) and ( b = 4 )- For ( x + frac{9}{x} = 6 ): [ x^2 - 6x + 9 = 0 implies (x - 3)^2 = 0 implies x = 3 ]- For ( y + frac{4}{y} = 4 ): [ y^2 - 4y + 4 = 0 implies (y - 2)^2 = 0 implies y = 2 ]Case 2: ( a = 4 ) and ( b = 6 )- For ( x + frac{9}{x} = 4 ): [ x^2 - 4x + 9 = 0 text{ (no real solutions)} ]- For ( y + frac{4}{y} = 6 ): [ y^2 - 6y + 4 = 0 text{ (real solutions, but } x text{ has no real solutions)} ]Only Case 1 yields real solutions. Verifying ( x = 3 ) and ( y = 2 ) in the original equations confirms they are correct. Negative values for ( x ) or ( y ) are not possible as they would result in negative terms which cannot satisfy the equations.Thus, the solution to the system is (boxed{(3, 2)}).]

🔑:The original system of equations can be transformed intoleft{begin{array}{l}left(x+frac{9}{x}right)+left(y+frac{4}{y}right)=10, left(x+frac{9}{x}right)left(y+frac{4}{y}right)=24 .end{array}right.By Vieta's Theorem, ( x+frac{9}{x} ) and ( y+frac{4}{y} ) are the roots of the equation ( t^{2}-10 t+24=0 ). Solving this equation, we get ( t=4 ) or ( t=6 ).thereforeleft{begin{array}{l}x+frac{9}{x}=4, y+frac{4}{y}=6end{array}right. text{ or } left{begin{array}{l}x+frac{9}{x}=6, y+frac{4}{y}=4 .end{array}right.Thus, (I) (left{begin{array}{l}x^{2}-4 x+9=0, y^{2}-6 y+4=0end{array}right.)or (II) (left{begin{array}{l}x^{2}-6 x+9=0, y^{2}-4 y+4=0 .end{array}right.)System (I) has no real solutions. Solving system (II), we getleft{begin{array}{l}x=3, y=2 .end{array}right.Upon verification, (left{begin{array}{l}x=3, y=2end{array}right.) is a solution to the original system of equations.