answer：First, I need to determine how much Alice has earned from selling each type of shoe.She sold 8 pairs of Nikes at 60 each, which amounts to 8 multiplied by 60, totaling 480.Next, she sold 6 pairs of Adidas shoes at 45 each, so 6 multiplied by 45 equals 270.Then, she sold 9 pairs of Reeboks at 35 each, resulting in 9 multiplied by 35, which is 315.Adding up all these amounts: 480 plus 270 plus 315 equals 1,065.Alice's goal was to sell 1,000 worth of shoes. Since she earned 1,065, she exceeded her goal by 65.

answer：Okay, so I need to evaluate the definite integral of (3x² - 1) from 1 to 2. Hmm, let me think about how to approach this. I remember that to solve definite integrals, I first need to find the antiderivative of the function and then apply the Fundamental Theorem of Calculus. Alright, so let's start by finding the antiderivative of 3x² - 1. I think the antiderivative of xⁿ is (xⁿ⁺¹)/(n+1), right? So for 3x², the antiderivative should be 3*(x³)/3, which simplifies to x³. That makes sense because the derivative of x³ is 3x², so that checks out.Now, what about the -1 term? The antiderivative of a constant, like 1, is just x, because the derivative of x is 1. So, the antiderivative of -1 should be -x. Putting those together, the antiderivative of the entire function 3x² - 1 is x³ - x. Let me write that down: F(x) = x³ - x. Now, according to the Fundamental Theorem of Calculus, the definite integral from a to b is F(b) - F(a). So in this case, I need to compute F(2) - F(1). Let me calculate F(2) first. Plugging in 2 into F(x): F(2) = (2)³ - 2 = 8 - 2 = 6.Okay, that seems straightforward. Now, let's compute F(1):F(1) = (1)³ - 1 = 1 - 1 = 0.Wait, so F(1) is 0? That's interesting. So then, the definite integral from 1 to 2 is F(2) - F(1) = 6 - 0 = 6.Hmm, let me double-check my calculations to make sure I didn't make a mistake. First, the antiderivative: - The integral of 3x² is indeed x³ because (d/dx)x³ = 3x².- The integral of -1 is -x because (d/dx)(-x) = -1.So, F(x) = x³ - x is correct.Calculating F(2):2³ is 8, minus 2 is 6. That's correct.Calculating F(1):1³ is 1, minus 1 is 0. That's also correct.So, subtracting F(1) from F(2) gives 6 - 0 = 6. Wait, is there another way to approach this integral? Maybe by expanding it or using another method? Let me try integrating term by term again just to be sure.The integral from 1 to 2 of (3x² - 1) dx can be split into two separate integrals:∫₁² 3x² dx - ∫₁² 1 dx.Calculating each integral separately:First integral: ∫3x² dx from 1 to 2.The antiderivative is x³, so evaluating from 1 to 2:2³ - 1³ = 8 - 1 = 7.Second integral: ∫1 dx from 1 to 2.The antiderivative is x, so evaluating from 1 to 2:2 - 1 = 1.Now, subtract the second integral from the first:7 - 1 = 6.Okay, so that's the same result as before. So, that confirms that the integral is indeed 6.Just to make sure, maybe I can approximate the integral using Riemann sums or another numerical method to see if it's close to 6. Let's try a simple left Riemann sum with a few rectangles.The interval from 1 to 2 is 1 unit long. Let's divide it into, say, two subintervals: [1, 1.5] and [1.5, 2]. So, each subinterval has a width of 0.5.For the left Riemann sum, we evaluate the function at the left endpoint of each subinterval.First subinterval [1, 1.5]: left endpoint is 1.f(1) = 3(1)² - 1 = 3 - 1 = 2.Second subinterval [1.5, 2]: left endpoint is 1.5.f(1.5) = 3*(1.5)² - 1 = 3*(2.25) - 1 = 6.75 - 1 = 5.75.So, the left Riemann sum is (2 + 5.75)*0.5 = 7.75*0.5 = 3.875.Hmm, that's quite a bit less than 6. Maybe using more subintervals would give a better approximation.Let's try with four subintervals, each of width 0.25.Subintervals: [1, 1.25], [1.25, 1.5], [1.5, 1.75], [1.75, 2].Left endpoints: 1, 1.25, 1.5, 1.75.Compute f at each:f(1) = 2 (as before).f(1.25) = 3*(1.25)² - 1 = 3*(1.5625) - 1 = 4.6875 - 1 = 3.6875.f(1.5) = 5.75 (as before).f(1.75) = 3*(1.75)² - 1 = 3*(3.0625) - 1 = 9.1875 - 1 = 8.1875.Sum these up: 2 + 3.6875 + 5.75 + 8.1875 = let's compute step by step.2 + 3.6875 = 5.68755.6875 + 5.75 = 11.437511.4375 + 8.1875 = 19.625Multiply by the width, 0.25: 19.625 * 0.25 = 4.90625.Hmm, that's closer to 6, but still a bit less. So, with four subintervals, the left Riemann sum is approximately 4.90625.If I take more subintervals, the approximation should get better. But since the exact value is 6, and the Riemann sums are approaching it, that gives me more confidence that 6 is the correct answer.Alternatively, maybe I can use the trapezoidal rule for a better approximation with fewer subintervals. Let's try that with two subintervals.Trapezoidal rule formula: (Δx/2)*(f(a) + 2f(a + Δx) + f(b)).Here, a = 1, b = 2, Δx = 0.5.Compute f(1) = 2, f(1.5) = 5.75, f(2) = 3*(2)² - 1 = 12 - 1 = 11.So, the trapezoidal sum is (0.5/2)*(2 + 2*5.75 + 11) = (0.25)*(2 + 11.5 + 11) = 0.25*(24.5) = 6.125.That's pretty close to 6. So, with just two subintervals, the trapezoidal rule gives 6.125, which is very close to the exact value of 6.This further confirms that my earlier calculation of 6 is correct.Wait, just to make sure, let me compute the exact integral again step by step.Given ∫₁² (3x² - 1) dx.First, find the antiderivative:∫(3x² - 1) dx = ∫3x² dx - ∫1 dx = x³ - x + C.So, F(x) = x³ - x.Compute F(2) = 2³ - 2 = 8 - 2 = 6.Compute F(1) = 1³ - 1 = 1 - 1 = 0.Thus, the definite integral is F(2) - F(1) = 6 - 0 = 6.Yep, that's consistent.I think I've verified it enough. I don't see any mistakes in my reasoning. The integral evaluates to 6.**Final Answer**The value of the definite integral is boxed{6}.

answer：Okay, so I have this limit problem to solve: the limit as x approaches infinity of (2x³ - 3x² + 5x + 7) divided by (3x³ + 4x² - x + 2). Hmm, I remember that when dealing with limits at infinity for rational functions, the degrees of the numerator and denominator play a big role. Let me try to recall the exact rules.First, both the numerator and the denominator are polynomials. The numerator is a cubic polynomial because the highest power of x is 3, and the denominator is also a cubic polynomial. So, both have the same degree, which is 3. I think when the degrees are equal, the limit as x approaches infinity is the ratio of the leading coefficients. Is that right?Let me double-check. If the degrees are the same, then yes, the limit is the coefficient of the highest degree term in the numerator divided by the coefficient of the highest degree term in the denominator. So in this case, the leading term in the numerator is 2x³, and in the denominator, it's 3x³. So, the limit should be 2/3. That seems straightforward.But wait, maybe I should go through the process step by step to make sure I'm not missing anything. Sometimes, even if the degrees are the same, there might be some nuances. Let me try dividing both the numerator and the denominator by the highest power of x in the denominator, which is x³. That should simplify things.So, if I divide each term in the numerator by x³, I get:(2x³/x³) - (3x²/x³) + (5x/x³) + (7/x³) = 2 - 3/x + 5/x² + 7/x³.Similarly, dividing each term in the denominator by x³ gives:(3x³/x³) + (4x²/x³) - (x/x³) + (2/x³) = 3 + 4/x - 1/x² + 2/x³.So now, the original expression becomes:[2 - 3/x + 5/x² + 7/x³] divided by [3 + 4/x - 1/x² + 2/x³].Now, as x approaches infinity, the terms with x in the denominator will approach zero. That is, 3/x, 5/x², 7/x³, 4/x, -1/x², and 2/x³ all go to zero. So, the expression simplifies to 2/3. Yep, that confirms my initial thought.Just to be thorough, let me consider another approach. Maybe using L’Hospital’s Rule? Wait, does that apply here? L’Hospital’s Rule is used when we have an indeterminate form like 0/0 or ∞/∞. In this case, as x approaches infinity, both the numerator and denominator approach infinity, so it is indeed an ∞/∞ form. So, maybe I can apply L’Hospital’s Rule.But hold on, L’Hospital’s Rule says that if we have a limit of the form f(x)/g(x) where both f(x) and g(x) approach 0 or both approach infinity, then the limit is equal to the limit of f’(x)/g’(x), provided that this latter limit exists. So, let me try taking the derivatives.The derivative of the numerator, f(x) = 2x³ - 3x² + 5x + 7, is f’(x) = 6x² - 6x + 5.The derivative of the denominator, g(x) = 3x³ + 4x² - x + 2, is g’(x) = 9x² + 8x - 1.So, applying L’Hospital’s Rule, the limit becomes:lim(x→∞) [6x² - 6x + 5] / [9x² + 8x - 1].Hmm, this is still a limit of the form ∞/∞, so maybe I can apply L’Hospital’s Rule again. Let me take the derivatives once more.The second derivative of the numerator, f''(x) = 12x - 6.The second derivative of the denominator, g''(x) = 18x + 8.So, applying L’Hospital’s Rule a second time, the limit becomes:lim(x→∞) [12x - 6] / [18x + 8].This is still ∞/∞, so I can apply L’Hospital’s Rule one more time.The third derivative of the numerator, f'''(x) = 12.The third derivative of the denominator, g'''(x) = 18.Now, applying L’Hospital’s Rule a third time, the limit becomes:lim(x→∞) 12 / 18, which simplifies to 2/3.So, even after applying L’Hospital’s Rule three times, I still end up with 2/3. That reinforces my confidence in the answer.Alternatively, another method is to factor out the highest power of x from both the numerator and the denominator. Let me try that as well.Starting with the numerator: 2x³ - 3x² + 5x + 7.Factor out x³: x³(2 - 3/x + 5/x² + 7/x³).Similarly, the denominator: 3x³ + 4x² - x + 2.Factor out x³: x³(3 + 4/x - 1/x² + 2/x³).So, the expression becomes:[x³(2 - 3/x + 5/x² + 7/x³)] / [x³(3 + 4/x - 1/x² + 2/x³)].The x³ terms cancel out, leaving:(2 - 3/x + 5/x² + 7/x³) / (3 + 4/x - 1/x² + 2/x³).Again, as x approaches infinity, the terms with 1/x, 1/x², and 1/x³ go to zero, so we're left with 2/3.So, whether I use the method of dividing numerator and denominator by x³, factoring out x³, or applying L’Hospital’s Rule multiple times, I consistently get the same result: 2/3.I think I'm pretty confident now that the limit is 2/3. I don't see any mistakes in my reasoning, and all the methods point to the same answer. Maybe just to visualize it, if I plug in a very large x, say x = 1000, into the original expression, I can approximate the value.Calculating numerator: 2*(1000)^3 - 3*(1000)^2 + 5*(1000) + 7 ≈ 2,000,000,000 - 3,000,000 + 5,000 + 7 ≈ approximately 1,997,005,007.Denominator: 3*(1000)^3 + 4*(1000)^2 - 1000 + 2 ≈ 3,000,000,000 + 4,000,000 - 1000 + 2 ≈ approximately 3,004,000,002.Dividing these two: 1,997,005,007 / 3,004,000,002 ≈ approximately 0.665, which is roughly 2/3 (since 2 divided by 3 is approximately 0.6667). So, that numerical approximation also supports the analytical result.Therefore, I can conclude that the limit is indeed 2/3.**Final Answer**The limit is boxed{dfrac{2}{3}}.

answer：First, I'll examine the three given lines:1. (3y - 2x = 1)2. (x + 2y = 2)3. (4x - 6y = 5)I notice that the third line, (4x - 6y = 5), can be simplified by dividing both sides by 2, resulting in (2x - 3y = 2.5). Comparing this with the first line, (3y - 2x = 1), I see that they are multiples of each other but not identical. This means the two lines are parallel and will never intersect.Next, I'll find the intersection points between the first and second lines, and between the second and third lines.For the intersection of the first and second lines:- Solve (3y - 2x = 1) and (x + 2y = 2) simultaneously.- From the second equation, express (x) in terms of (y): (x = 2 - 2y).- Substitute this into the first equation: (3y - 2(2 - 2y) = 1).- Simplify and solve for (y): (3y - 4 + 4y = 1) → (7y = 5) → (y = frac{5}{7}).- Substitute (y = frac{5}{7}) back into (x = 2 - 2y) to find (x = frac{4}{7}).For the intersection of the second and third lines:- Solve (x + 2y = 2) and (4x - 6y = 5) simultaneously.- From the first equation, express (x) in terms of (y): (x = 2 - 2y).- Substitute this into the third equation: (4(2 - 2y) - 6y = 5).- Simplify and solve for (y): (8 - 8y - 6y = 5) → (-14y = -3) → (y = frac{3}{14}).- Substitute (y = frac{3}{14}) back into (x = 2 - 2y) to find (x = frac{11}{7}).Since the first and third lines are parallel and do not intersect, there are no additional intersection points. Therefore, there are two distinct points where at least two of the three lines intersect.

answer：Okay, so I have this problem here about an arithmetic sequence. Let me try to understand what it's asking. It says that {a_n} is an arithmetic sequence, and S_n is the sum of its first n terms. They give me that a_1 is 6, and that a_3 + a_5 equals 0. I need to find S_6, which is the sum of the first six terms. Hmm, okay.First, let me recall what an arithmetic sequence is. An arithmetic sequence is a sequence where each term after the first is obtained by adding a constant difference. So, if the first term is a_1, then the second term is a_1 + d, the third term is a_1 + 2d, and so on. So, in general, the nth term a_n can be written as a_n = a_1 + (n - 1)d, where d is the common difference.Given that a_1 is 6, so that's straightforward. Now, they also tell me that a_3 + a_5 = 0. Let me write expressions for a_3 and a_5 using the formula for the nth term.So, a_3 = a_1 + 2d, because it's the third term. Similarly, a_5 = a_1 + 4d. Adding these together, a_3 + a_5 = (a_1 + 2d) + (a_1 + 4d) = 2a_1 + 6d. They say this equals 0. So, 2a_1 + 6d = 0.Since a_1 is 6, let me substitute that in. So, 2*6 + 6d = 0. That simplifies to 12 + 6d = 0. Now, solving for d, I can subtract 12 from both sides: 6d = -12. Then, divide both sides by 6: d = -2.Okay, so the common difference d is -2. That means each term is 2 less than the previous term.Now, I need to find S_6, the sum of the first six terms. I remember that the formula for the sum of the first n terms of an arithmetic sequence is S_n = n/2 * (2a_1 + (n - 1)d). Alternatively, it can also be written as S_n = n*(a_1 + a_n)/2, since the average of the first and nth term multiplied by the number of terms gives the sum.Let me use the first formula: S_n = n/2 * (2a_1 + (n - 1)d). So, plugging in n = 6, a_1 = 6, and d = -2.First, calculate 2a_1: 2*6 = 12.Then, calculate (n - 1)d: (6 - 1)*(-2) = 5*(-2) = -10.Add those together: 12 + (-10) = 2.Now, multiply by n/2: 6/2 * 2 = 3*2 = 6.Wait, so S_6 is 6? Hmm, that seems a bit low. Let me double-check my calculations.Alternatively, I can use the other formula: S_n = n*(a_1 + a_n)/2. For this, I need to find a_6 first.a_6 = a_1 + 5d = 6 + 5*(-2) = 6 - 10 = -4.So, a_6 is -4. Then, S_6 = 6*(6 + (-4))/2 = 6*(2)/2 = 6*1 = 6. Hmm, same result.Wait, so that's correct? Let me list out the terms to make sure.a_1 = 6a_2 = a_1 + d = 6 + (-2) = 4a_3 = a_2 + d = 4 + (-2) = 2a_4 = a_3 + d = 2 + (-2) = 0a_5 = a_4 + d = 0 + (-2) = -2a_6 = a_5 + d = -2 + (-2) = -4So, the first six terms are: 6, 4, 2, 0, -2, -4.Adding them up: 6 + 4 = 10; 10 + 2 = 12; 12 + 0 = 12; 12 + (-2) = 10; 10 + (-4) = 6. Yeah, that adds up to 6. So, S_6 is indeed 6.Wait, but just to make sure, let me try another approach. Maybe using the formula for the sum with the average of the first and last term.So, S_6 = 6*(a_1 + a_6)/2 = 6*(6 + (-4))/2 = 6*(2)/2 = 6*1 = 6. Yep, same answer.Alternatively, I can think about the sequence and see if it makes sense. The terms are decreasing by 2 each time, starting from 6. So, the terms are 6, 4, 2, 0, -2, -4. If I plot these on a number line, they are symmetric around 1. Wait, is that true?Wait, 6 and -4: the average is (6 + (-4))/2 = 1.Similarly, 4 and -2: average is (4 + (-2))/2 = 1.And 2 and 0: average is (2 + 0)/2 = 1.So, each pair adds up to 2, and there are three pairs. So, 3*2 = 6. That's another way to see it.So, yeah, that makes sense. So, S_6 is 6.I think I'm confident now that the answer is 6.**Final Answer**boxed{6}

answer：Alright, so I have this problem here: I need to find the smallest acute angle θ such that sin θ, sin 2θ, and sin 3θ form an arithmetic progression in some order. Then, I have to find cos θ. Hmm, okay. Let me try to break this down step by step.First, let me recall what an arithmetic progression (AP) is. It's a sequence where the difference between consecutive terms is constant. So, if I have three terms a, b, c in AP, then 2b = a + c. That means the middle term is the average of the other two.Now, the problem says that sin θ, sin 2θ, and sin 3θ form an AP in some order. That means these three terms can be arranged in some order such that the middle one is the average of the other two. So, there are a few possibilities here because the order isn't specified. I need to consider all possible orderings and see which one gives me a valid θ.Let me list all the possible orderings of the three terms:1. sin θ, sin 2θ, sin 3θ2. sin θ, sin 3θ, sin 2θ3. sin 2θ, sin θ, sin 3θ4. sin 2θ, sin 3θ, sin θ5. sin 3θ, sin θ, sin 2θ6. sin 3θ, sin 2θ, sin θFor each of these orderings, I can set up the condition for an arithmetic progression and solve for θ. Then, among all possible solutions, I need to find the smallest acute angle θ.But before diving into all six cases, maybe I can find a smarter approach. Perhaps using some trigonometric identities or properties to simplify the problem.I remember that sin 3θ can be expressed in terms of sin θ and sin 2θ. Let me recall the triple-angle identity:sin 3θ = 3 sin θ - 4 sin³ θ.Hmm, that might come in handy. Alternatively, I can use the identity for sin 3θ in terms of sin θ and cos θ:sin 3θ = 3 sin θ - 4 sin³ θ.But maybe that's more complicated. Alternatively, I can use the identity:sin 3θ = sin(2θ + θ) = sin 2θ cos θ + cos 2θ sin θ.Yes, that might be useful. Let me write that down:sin 3θ = sin 2θ cos θ + cos 2θ sin θ.Similarly, I know that sin 2θ = 2 sin θ cos θ.So, maybe I can express everything in terms of sin θ and cos θ, and then set up equations based on the AP condition.Alternatively, perhaps I can consider the differences between the terms. Since in an AP, the difference between consecutive terms is constant, so for any ordering, the difference between the first and second term should be equal to the difference between the second and third term.Let me consider each case one by one.Case 1: sin θ, sin 2θ, sin 3θ are in AP.So, the condition is 2 sin 2θ = sin θ + sin 3θ.Let me compute sin 3θ using the identity:sin 3θ = 3 sin θ - 4 sin³ θ.So, substituting into the equation:2 sin 2θ = sin θ + 3 sin θ - 4 sin³ θSimplify the right-hand side:2 sin 2θ = 4 sin θ - 4 sin³ θDivide both sides by 2:sin 2θ = 2 sin θ - 2 sin³ θBut sin 2θ is equal to 2 sin θ cos θ, so:2 sin θ cos θ = 2 sin θ - 2 sin³ θDivide both sides by 2 sin θ (assuming sin θ ≠ 0, which it isn't since θ is acute):cos θ = 1 - sin² θBut 1 - sin² θ is equal to cos² θ, so:cos θ = cos² θBring everything to one side:cos² θ - cos θ = 0Factor:cos θ (cos θ - 1) = 0So, cos θ = 0 or cos θ = 1.But θ is an acute angle, so cos θ = 0 would mean θ = π/2, which is 90 degrees, and cos θ = 1 would mean θ = 0, which isn't an acute angle in the sense of being between 0 and π/2, exclusive. So, θ = π/2 is a possibility, but let's check if that works.Wait, if θ = π/2, then sin θ = 1, sin 2θ = sin π = 0, sin 3θ = sin (3π/2) = -1. So, the terms are 1, 0, -1, which is indeed an arithmetic progression with common difference -1. But θ = π/2 is the largest acute angle, not the smallest. The problem asks for the smallest acute angle. So, maybe this case doesn't give us the smallest θ. Let's move on to the next case.Case 2: sin θ, sin 3θ, sin 2θ are in AP.So, the condition is 2 sin 3θ = sin θ + sin 2θ.Again, let's use the identity for sin 3θ:sin 3θ = 3 sin θ - 4 sin³ θ.So, substituting:2(3 sin θ - 4 sin³ θ) = sin θ + 2 sin θ cos θSimplify left side:6 sin θ - 8 sin³ θ = sin θ + 2 sin θ cos θBring all terms to left side:6 sin θ - 8 sin³ θ - sin θ - 2 sin θ cos θ = 0Simplify:5 sin θ - 8 sin³ θ - 2 sin θ cos θ = 0Factor out sin θ:sin θ (5 - 8 sin² θ - 2 cos θ) = 0Again, sin θ ≠ 0, so:5 - 8 sin² θ - 2 cos θ = 0But sin² θ = 1 - cos² θ, so substitute:5 - 8(1 - cos² θ) - 2 cos θ = 0Simplify:5 - 8 + 8 cos² θ - 2 cos θ = 0Which is:-3 + 8 cos² θ - 2 cos θ = 0Multiply both sides by -1:3 - 8 cos² θ + 2 cos θ = 0Rearrange:8 cos² θ - 2 cos θ - 3 = 0Now, this is a quadratic in cos θ. Let me denote x = cos θ:8x² - 2x - 3 = 0Solve for x:Using quadratic formula:x = [2 ± sqrt(4 + 96)] / 16Because discriminant D = ( -2 )² - 4*8*(-3) = 4 + 96 = 100So,x = [2 ± 10] / 16So, two solutions:x = (2 + 10)/16 = 12/16 = 3/4x = (2 - 10)/16 = (-8)/16 = -1/2So, cos θ = 3/4 or cos θ = -1/2.But θ is acute, so cos θ must be positive. So, cos θ = 3/4.So, θ = arccos(3/4). Let me compute this angle.But before I get too excited, I need to check if this θ actually satisfies the original condition.So, let's compute sin θ, sin 2θ, sin 3θ and see if they form an AP in the order sin θ, sin 3θ, sin 2θ.First, cos θ = 3/4, so sin θ = sqrt(1 - (9/16)) = sqrt(7/16) = sqrt7 / 4.Compute sin θ = sqrt7 / 4 ≈ 0.6614Compute sin 2θ = 2 sin θ cos θ = 2*(sqrt7 / 4)*(3/4) = (6 sqrt7)/16 ≈ 0.6614*1.5 ≈ 0.9921Wait, let me compute that more accurately:sqrt7 ≈ 2.6458, so sqrt7 / 4 ≈ 0.6614So, sin 2θ = 2*(0.6614)*(0.75) = 2*0.6614*0.75 ≈ 0.9921Compute sin 3θ: Let's use the identity sin 3θ = 3 sin θ - 4 sin³ θ.So, sin 3θ = 3*(sqrt7 / 4) - 4*(sqrt7 / 4)^3Compute each term:First term: 3*(sqrt7 / 4) ≈ 3*0.6614 ≈ 1.9842Second term: 4*( (sqrt7)^3 ) / (4^3 ) = 4*(7*sqrt7)/64 = (28 sqrt7)/64 = (7 sqrt7)/16 ≈ (7*2.6458)/16 ≈ 18.5206 /16 ≈ 1.1575So, sin 3θ ≈ 1.9842 - 1.1575 ≈ 0.8267Wait, so sin θ ≈ 0.6614, sin 3θ ≈ 0.8267, sin 2θ ≈ 0.9921So, the order is sin θ ≈ 0.6614, sin 3θ ≈ 0.8267, sin 2θ ≈ 0.9921Is this an arithmetic progression? Let's check the differences:0.8267 - 0.6614 ≈ 0.16530.9921 - 0.8267 ≈ 0.1654So, approximately equal differences. So, yes, it's an AP with common difference ≈0.1653.So, this works. So, cos θ = 3/4 is a solution in this case.But before I conclude, let me check the other cases to see if there's a smaller θ.Case 3: sin 2θ, sin θ, sin 3θ are in AP.So, the condition is 2 sin θ = sin 2θ + sin 3θ.Let me write that:2 sin θ = sin 2θ + sin 3θExpress sin 3θ as before:sin 3θ = 3 sin θ - 4 sin³ θSo,2 sin θ = sin 2θ + 3 sin θ - 4 sin³ θSimplify:2 sin θ = 2 sin θ cos θ + 3 sin θ - 4 sin³ θBring all terms to left:2 sin θ - 2 sin θ cos θ - 3 sin θ + 4 sin³ θ = 0Simplify:(- sin θ) - 2 sin θ cos θ + 4 sin³ θ = 0Factor out sin θ:sin θ (-1 - 2 cos θ + 4 sin² θ) = 0Again, sin θ ≠ 0, so:-1 - 2 cos θ + 4 sin² θ = 0Express sin² θ as 1 - cos² θ:-1 - 2 cos θ + 4(1 - cos² θ) = 0Simplify:-1 - 2 cos θ + 4 - 4 cos² θ = 0Which is:3 - 2 cos θ - 4 cos² θ = 0Rearrange:-4 cos² θ - 2 cos θ + 3 = 0Multiply both sides by -1:4 cos² θ + 2 cos θ - 3 = 0Quadratic in cos θ: 4x² + 2x - 3 = 0Solutions:x = [-2 ± sqrt(4 + 48)] / 8 = [-2 ± sqrt(52)] / 8 = [-2 ± 2 sqrt13] /8 = [-1 ± sqrt13]/4So, two solutions:x = (-1 + sqrt13)/4 ≈ (-1 + 3.6055)/4 ≈ 2.6055/4 ≈ 0.6514x = (-1 - sqrt13)/4 ≈ negative value, which we can ignore since θ is acute.So, cos θ ≈ 0.6514, which is approximately 49 degrees (since cos 49 ≈ 0.656). Wait, but 0.6514 is slightly less, so maybe around 49 degrees.But let's compute it more accurately:sqrt13 ≈ 3.6055, so (-1 + 3.6055)/4 ≈ 2.6055 /4 ≈ 0.6514So, θ ≈ arccos(0.6514) ≈ 49 degrees.Wait, earlier in Case 2, we had cos θ = 3/4 ≈ 0.75, which is about 41 degrees.So, 41 degrees is smaller than 49 degrees, so 41 degrees is a smaller angle. So, Case 2 gives a smaller θ.But let me check if this solution in Case 3 actually satisfies the original condition.Compute sin 2θ, sin θ, sin 3θ.Given cos θ ≈ 0.6514, so sin θ ≈ sqrt(1 - 0.6514²) ≈ sqrt(1 - 0.4243) ≈ sqrt(0.5757) ≈ 0.7587Compute sin 2θ = 2 sin θ cos θ ≈ 2*0.7587*0.6514 ≈ 1.5174*0.6514 ≈ 0.988Compute sin 3θ: Using sin 3θ = 3 sin θ - 4 sin³ θSo, sin 3θ ≈ 3*0.7587 - 4*(0.7587)^3 ≈ 2.2761 - 4*(0.436) ≈ 2.2761 - 1.744 ≈ 0.5321So, the terms are:sin 2θ ≈ 0.988, sin θ ≈ 0.7587, sin 3θ ≈ 0.5321Wait, the order is sin 2θ, sin θ, sin 3θ, so 0.988, 0.7587, 0.5321Is this an AP? Let's check the differences:0.7587 - 0.988 ≈ -0.22930.5321 - 0.7587 ≈ -0.2266These are approximately equal, so yes, it's an AP with common difference ≈ -0.22795So, this also works. So, in this case, cos θ ≈ 0.6514, which is approximately 49 degrees, which is larger than the 41 degrees from Case 2. So, Case 2 gives a smaller θ.Case 4: sin 2θ, sin 3θ, sin θ are in AP.So, the condition is 2 sin 3θ = sin 2θ + sin θ.Again, let's use the identity for sin 3θ:sin 3θ = 3 sin θ - 4 sin³ θSo,2(3 sin θ - 4 sin³ θ) = sin 2θ + sin θSimplify left side:6 sin θ - 8 sin³ θ = sin 2θ + sin θBring all terms to left:6 sin θ - 8 sin³ θ - sin 2θ - sin θ = 0Simplify:5 sin θ - 8 sin³ θ - sin 2θ = 0Express sin 2θ as 2 sin θ cos θ:5 sin θ - 8 sin³ θ - 2 sin θ cos θ = 0Factor out sin θ:sin θ (5 - 8 sin² θ - 2 cos θ) = 0Again, sin θ ≠ 0, so:5 - 8 sin² θ - 2 cos θ = 0Express sin² θ as 1 - cos² θ:5 - 8(1 - cos² θ) - 2 cos θ = 0Simplify:5 - 8 + 8 cos² θ - 2 cos θ = 0Which is:-3 + 8 cos² θ - 2 cos θ = 0Multiply by -1:3 - 8 cos² θ + 2 cos θ = 0Rearrange:8 cos² θ - 2 cos θ - 3 = 0Wait, this is the same quadratic as in Case 2:8x² - 2x - 3 = 0, where x = cos θSo, solutions are x = 3/4 and x = -1/2, as before.Since θ is acute, cos θ = 3/4, which is the same solution as in Case 2.So, this case doesn't give a new solution, just the same θ as Case 2.Case 5: sin 3θ, sin θ, sin 2θ are in AP.So, the condition is 2 sin θ = sin 3θ + sin 2θ.Let me write that:2 sin θ = sin 3θ + sin 2θExpress sin 3θ as before:sin 3θ = 3 sin θ - 4 sin³ θSo,2 sin θ = 3 sin θ - 4 sin³ θ + sin 2θBring all terms to left:2 sin θ - 3 sin θ + 4 sin³ θ - sin 2θ = 0Simplify:- sin θ + 4 sin³ θ - sin 2θ = 0Express sin 2θ as 2 sin θ cos θ:- sin θ + 4 sin³ θ - 2 sin θ cos θ = 0Factor out sin θ:sin θ (-1 + 4 sin² θ - 2 cos θ) = 0Again, sin θ ≠ 0, so:-1 + 4 sin² θ - 2 cos θ = 0Express sin² θ as 1 - cos² θ:-1 + 4(1 - cos² θ) - 2 cos θ = 0Simplify:-1 + 4 - 4 cos² θ - 2 cos θ = 0Which is:3 - 4 cos² θ - 2 cos θ = 0Rearrange:-4 cos² θ - 2 cos θ + 3 = 0Multiply by -1:4 cos² θ + 2 cos θ - 3 = 0Quadratic in cos θ: 4x² + 2x - 3 = 0Solutions:x = [-2 ± sqrt(4 + 48)] / 8 = [-2 ± sqrt(52)] / 8 = [-2 ± 2 sqrt13]/8 = [-1 ± sqrt13]/4So, x = (-1 + sqrt13)/4 ≈ 0.6514 or x = (-1 - sqrt13)/4 (negative, discard)So, cos θ ≈ 0.6514, which is the same as in Case 3.So, this case gives the same solution as Case 3.Case 6: sin 3θ, sin 2θ, sin θ are in AP.So, the condition is 2 sin 2θ = sin 3θ + sin θ.Let me write that:2 sin 2θ = sin 3θ + sin θExpress sin 3θ as before:sin 3θ = 3 sin θ - 4 sin³ θSo,2 sin 2θ = 3 sin θ - 4 sin³ θ + sin θSimplify right side:2 sin 2θ = 4 sin θ - 4 sin³ θExpress sin 2θ as 2 sin θ cos θ:2*(2 sin θ cos θ) = 4 sin θ - 4 sin³ θSimplify left side:4 sin θ cos θ = 4 sin θ - 4 sin³ θDivide both sides by 4 sin θ (assuming sin θ ≠ 0):cos θ = 1 - sin² θBut 1 - sin² θ = cos² θ, so:cos θ = cos² θWhich gives:cos² θ - cos θ = 0Factor:cos θ (cos θ - 1) = 0So, cos θ = 0 or cos θ = 1.Again, θ is acute, so cos θ = 0 would mean θ = π/2, which is 90 degrees, and cos θ = 1 would mean θ = 0, which isn't an acute angle. So, θ = π/2 is the solution here, which is the same as in Case 1.So, this case doesn't give a new solution.So, summarizing all cases:- Case 1 and 6 give θ = π/2 (90 degrees), which is the largest acute angle.- Case 2 and 4 give cos θ = 3/4, θ ≈ 41 degrees.- Case 3 and 5 give cos θ ≈ 0.6514, θ ≈ 49 degrees.So, the smallest acute angle θ is the one with cos θ = 3/4, which is approximately 41 degrees.Therefore, the answer is cos θ = 3/4.But just to be thorough, let me check if there are any other possible cases or if I missed something.Wait, in Case 2, we had cos θ = 3/4, which is approximately 0.75, which is a valid acute angle. The other solution in Case 3 was approximately 0.6514, which is about 49 degrees, which is larger than 41 degrees, so 41 degrees is indeed the smallest.Therefore, the smallest acute angle θ is arccos(3/4), so cos θ = 3/4.**Final Answer**The value of cos theta is boxed{dfrac{3}{4}}.