💡Alright, so I have this problem where I need to find the complex numbers α and β such that the equation |z̄|² + αz + ᾱz̄ + β = 0 represents the circumcircle of triangle ABC. The vertices of triangle ABC are given as complex numbers: 3 + 2i, 3i, and 2 - i. First, I need to recall what the circumcircle of a triangle is. It's the unique circle that passes through all three vertices of the triangle. So, my goal is to find the equation of this circle in the complex plane and then express it in the given form to identify α and β.The equation given is |z̄|² + αz + ᾱz̄ + β = 0. I know that |z̄|² is the same as |z|² because the modulus squared of a complex number is equal to the modulus squared of its conjugate. So, |z̄|² = |z|². Therefore, the equation simplifies to |z|² + αz + ᾱz̄ + β = 0.I remember that in the complex plane, the general equation of a circle can be written as |z|² + αz + ᾱz̄ + β = 0, where α is a complex constant and β is a real constant. So, this is the standard form I need to work with.Now, I need to find the circumcircle of triangle ABC. To do this, I can use the coordinates of the vertices. Let me convert the complex numbers to their coordinate forms:- A: 3 + 2i corresponds to the point (3, 2)- B: 3i corresponds to the point (0, 3)- C: 2 - i corresponds to the point (2, -1)So, the triangle has vertices at (3, 2), (0, 3), and (2, -1). I need to find the equation of the circle passing through these three points.I recall that the general equation of a circle in the Cartesian plane is x² + y² + Dx + Ey + F = 0, where D, E, and F are constants. Since the given equation in complex form is |z|² + αz + ᾱz̄ + β = 0, I can relate this to the Cartesian form.Let me write z as x + yi, where x and y are real numbers. Then, |z|² = x² + y², αz = α(x + yi), and ᾱz̄ = ᾱ(x - yi). So, expanding the equation:|z|² + αz + ᾱz̄ + β = x² + y² + α(x + yi) + ᾱ(x - yi) + β = 0.Let me denote α as a + bi, where a and b are real numbers. Then, ᾱ is a - bi. Substituting these into the equation:x² + y² + (a + bi)(x + yi) + (a - bi)(x - yi) + β = 0.Let me expand the terms:First, expand (a + bi)(x + yi):= a*x + a*yi + bi*x + bi*yi= ax + ayi + bxi + b y i²= ax + (ay + bx)i + b y (-1)= ax - by + (ay + bx)i.Similarly, expand (a - bi)(x - yi):= a*x - a*yi - bi*x + bi*yi= ax - ayi - bxi + b y i²= ax - ayi - bxi + b y (-1)= ax - by - (ay + bx)i.Now, add these two results together:(ax - by + (ay + bx)i) + (ax - by - (ay + bx)i) = 2ax - 2by.So, the imaginary parts cancel out, and we are left with 2ax - 2by.Therefore, the entire equation becomes:x² + y² + 2ax - 2by + β = 0.Comparing this to the general Cartesian equation x² + y² + Dx + Ey + F = 0, we can see that:D = 2a,E = -2b,F = β.So, if I can find D, E, and F for the circumcircle of triangle ABC, I can solve for a, b, and β.To find D, E, and F, I need to plug in the coordinates of points A, B, and C into the general circle equation and solve the system of equations.Let's write down the equations for each point:For point A (3, 2):3² + 2² + D*3 + E*2 + F = 09 + 4 + 3D + 2E + F = 013 + 3D + 2E + F = 0 ...(1)For point B (0, 3):0² + 3² + D*0 + E*3 + F = 00 + 9 + 0 + 3E + F = 09 + 3E + F = 0 ...(2)For point C (2, -1):2² + (-1)² + D*2 + E*(-1) + F = 04 + 1 + 2D - E + F = 05 + 2D - E + F = 0 ...(3)Now, I have three equations:1) 13 + 3D + 2E + F = 02) 9 + 3E + F = 03) 5 + 2D - E + F = 0I need to solve this system of equations for D, E, and F.Let me subtract equation (2) from equation (1):(13 + 3D + 2E + F) - (9 + 3E + F) = 0 - 013 - 9 + 3D - 0 + 2E - 3E + F - F = 04 + 3D - E = 0So, 3D - E = -4 ...(4)Similarly, subtract equation (2) from equation (3):(5 + 2D - E + F) - (9 + 3E + F) = 0 - 05 - 9 + 2D - 0 - E - 3E + F - F = 0-4 + 2D - 4E = 0So, 2D - 4E = 4 ...(5)Now, I have equations (4) and (5):(4): 3D - E = -4(5): 2D - 4E = 4I can solve this system for D and E.First, let me solve equation (4) for E:E = 3D + 4Now, substitute E into equation (5):2D - 4*(3D + 4) = 42D - 12D - 16 = 4-10D - 16 = 4-10D = 20D = -2Now, substitute D = -2 into equation (4):3*(-2) - E = -4-6 - E = -4-E = 2E = -2Now, substitute D = -2 and E = -2 into equation (2):9 + 3*(-2) + F = 09 - 6 + F = 03 + F = 0F = -3So, we have D = -2, E = -2, F = -3.Now, recall that D = 2a, E = -2b, and F = β.So, D = 2a = -2 => a = -1E = -2b = -2 => b = 1F = β = -3Therefore, α = a + bi = -1 + iAnd β = -3.So, the complex numbers are α = -1 + i and β = -3.Let me double-check my work to ensure there are no mistakes.First, I found the coordinates of the points correctly.Then, I set up the general equation of the circle and expanded the given complex equation correctly, leading to x² + y² + 2ax - 2by + β = 0.I correctly identified D = 2a, E = -2b, F = β.Then, I plugged in the coordinates into the general circle equation, resulting in three equations.Solving the system, I found D = -2, E = -2, F = -3.Then, using D = 2a, E = -2b, I found a = -1, b = 1, so α = -1 + i, and β = -3.To ensure this is correct, I can plug the values back into the original equation and verify if the points lie on the circle.Let me check point A (3, 2):|z|² + αz + ᾱz̄ + βz = 3 + 2i, so |z|² = 3² + 2² = 9 + 4 = 13αz = (-1 + i)(3 + 2i) = (-1)(3) + (-1)(2i) + i(3) + i(2i) = -3 - 2i + 3i + 2i² = -3 + i + 2(-1) = -3 + i - 2 = -5 + iᾱz̄ = (-1 - i)(3 - 2i) = (-1)(3) + (-1)(-2i) + (-i)(3) + (-i)(-2i) = -3 + 2i - 3i + 2i² = -3 - i + 2(-1) = -3 - i - 2 = -5 - iAdding these together:|z|² + αz + ᾱz̄ + β = 13 + (-5 + i) + (-5 - i) + (-3) = 13 -5 -5 -3 + (i - i) = 0 + 0i = 0So, point A satisfies the equation.Similarly, check point B (0, 3):z = 0 + 3i, |z|² = 0² + 3² = 9αz = (-1 + i)(0 + 3i) = (-1)(0) + (-1)(3i) + i(0) + i(3i) = 0 - 3i + 0 + 3i² = -3i + 3(-1) = -3i - 3ᾱz̄ = (-1 - i)(0 - 3i) = (-1)(0) + (-1)(-3i) + (-i)(0) + (-i)(-3i) = 0 + 3i + 0 + 3i² = 3i + 3(-1) = 3i - 3Adding these together:|z|² + αz + ᾱz̄ + β = 9 + (-3i - 3) + (3i - 3) + (-3) = 9 -3 -3 -3 + (-3i + 3i) = 0 + 0i = 0Point B satisfies the equation.Finally, check point C (2, -1):z = 2 - i, |z|² = 2² + (-1)² = 4 + 1 = 5αz = (-1 + i)(2 - i) = (-1)(2) + (-1)(-i) + i(2) + i(-i) = -2 + i + 2i - i² = -2 + 3i - (-1) = -2 + 3i + 1 = -1 + 3iᾱz̄ = (-1 - i)(2 + i) = (-1)(2) + (-1)(i) + (-i)(2) + (-i)(i) = -2 - i - 2i - i² = -2 - 3i - (-1) = -2 - 3i + 1 = -1 - 3iAdding these together:|z|² + αz + ᾱz̄ + β = 5 + (-1 + 3i) + (-1 - 3i) + (-3) = 5 -1 -1 -3 + (3i - 3i) = 0 + 0i = 0Point C also satisfies the equation.Therefore, my solution is correct. The complex numbers α and β are α = -1 + i and β = -3.**Final Answer**The complex numbers are (boxed{alpha = -1 + i}) and (boxed{beta = -3}).

💡First, I need to determine the number of members in each group. Let's denote the number of members in the second group as ( x ). According to the problem, the first group has 5 more members than the second group, so the number of members in the first group will be ( x + 5 ).Each member of the second group contributes ( x ) paise, so the total collection from the second group will be ( x times x ) paise. Similarly, each member of the first group contributes ( (x + 5) ) paise, so the total collection from the first group will be ( (x + 5) times (x + 5) ) paise.The total collection from both groups is Rs. 130, which is equivalent to ( 130 times 100 = 13000 ) paise (since 1 Rupee = 100 paise).So, I can set up the following equation:[x^2 + (x + 5)^2 = 13000]Expanding the equation, I get:[x^2 + x^2 + 10x + 25 = 13000]Combining like terms, I get:[2x^2 + 10x + 25 = 13000]Subtracting 13000 from both sides, I get:[2x^2 + 10x - 12975 = 0]Now, I need to solve this quadratic equation for ( x ). I'll use the quadratic formula:[x = frac{-b pm sqrt{b^2 - 4ac}}{2a}]where ( a = 2 ), ( b = 10 ), and ( c = -12975 ).First, I'll calculate the discriminant:[text{Discriminant} = 10^2 - 4 times 2 times (-12975) = 100 + 103800 = 103900]Since the discriminant is positive, there are two real solutions. Now, I'll find the values of ( x ):[x = frac{-10 pm sqrt{103900}}{4}]Calculating the square root of 103900 gives approximately 322. So,[x_1 = frac{-10 + 322}{4} = frac{312}{4} = 78][x_2 = frac{-10 - 322}{4} = frac{-332}{4} = -83]Since the number of members cannot be negative, I discard ( x_2 = -83 ).Therefore, the number of members in the second group is 78, and the number of members in the first group is ( 78 + 5 = 83 ).

💡Okay, so I have this problem where I'm given that the tangent of (α + π/4) is equal to -3, and α is between 0 and π/2. I need to find the value of tan α and then use that to find sin(2α - π/3). Hmm, let's start with part (1).First, I remember that there's a tangent addition formula. It goes like this: tan(A + B) = (tan A + tan B) / (1 - tan A tan B). So, in this case, A is α and B is π/4. I know that tan(π/4) is 1, so plugging that into the formula, I get tan(α + π/4) = (tan α + 1) / (1 - tan α). The problem says that this equals -3. So, I can set up the equation: (tan α + 1) / (1 - tan α) = -3. Now, I need to solve for tan α. Let me write that equation down:(tan α + 1) / (1 - tan α) = -3To solve for tan α, I can cross-multiply. So, multiplying both sides by (1 - tan α) gives:tan α + 1 = -3(1 - tan α)Let me expand the right side:tan α + 1 = -3 + 3 tan αNow, I want to get all the tan α terms on one side and constants on the other. So, subtract tan α from both sides:1 = -3 + 2 tan αThen, add 3 to both sides:4 = 2 tan αDivide both sides by 2:tan α = 2Wait, that seems straightforward. So, tan α is 2. But let me double-check because sometimes with trigonometric identities, especially when dealing with quadrants, things can get tricky.Given that α is between 0 and π/2, which is the first quadrant, all trigonometric functions should be positive there. So, tan α is positive, which makes sense because we got 2, a positive number. So, that seems consistent.Okay, so part (1) is done. Now, moving on to part (2): finding sin(2α - π/3). Hmm, this seems a bit more involved. I think I can use the sine of a difference formula here, which is sin(A - B) = sin A cos B - cos A sin B. So, in this case, A is 2α and B is π/3.So, sin(2α - π/3) = sin(2α) cos(π/3) - cos(2α) sin(π/3). I know the values of cos(π/3) and sin(π/3). Cos(π/3) is 0.5 and sin(π/3) is (√3)/2. So, plugging those in:sin(2α - π/3) = sin(2α) * 0.5 - cos(2α) * (√3)/2Now, I need to find sin(2α) and cos(2α). I remember that there are double-angle formulas for sine and cosine. For sine, sin(2α) = 2 sin α cos α, and for cosine, cos(2α) = cos² α - sin² α. Alternatively, cos(2α) can also be written as 1 - 2 sin² α or 2 cos² α - 1. Maybe one of those forms will be easier to use given that I know tan α.Since I know tan α is 2, I can find sin α and cos α. Let me think. If tan α = 2, that means in a right triangle, the opposite side is 2 and the adjacent side is 1, so the hypotenuse would be sqrt(1² + 2²) = sqrt(5). Therefore, sin α is opposite over hypotenuse, which is 2/sqrt(5), and cos α is adjacent over hypotenuse, which is 1/sqrt(5).So, sin α = 2/√5 and cos α = 1/√5.Now, let's compute sin(2α):sin(2α) = 2 sin α cos α = 2*(2/√5)*(1/√5) = 2*(2/5) = 4/5Okay, so sin(2α) is 4/5.Now, let's compute cos(2α). Using the identity cos(2α) = cos² α - sin² α:cos(2α) = (1/√5)² - (2/√5)² = (1/5) - (4/5) = -3/5So, cos(2α) is -3/5.Wait, that's interesting. So, even though α is in the first quadrant, 2α could be in the second quadrant because α is less than π/2, so 2α is less than π. So, cos(2α) is negative, which makes sense because in the second quadrant, cosine is negative.Okay, so now I have sin(2α) = 4/5 and cos(2α) = -3/5.Going back to the expression for sin(2α - π/3):sin(2α - π/3) = sin(2α)*0.5 - cos(2α)*(√3)/2Plugging in the values:= (4/5)*(1/2) - (-3/5)*(√3/2)Let me compute each term separately.First term: (4/5)*(1/2) = 4/10 = 2/5Second term: (-3/5)*(√3/2) = (-3√3)/10But since it's minus this term, it becomes:= 2/5 - (-3√3)/10 = 2/5 + 3√3/10To combine these, I need a common denominator. The denominators are 5 and 10, so 10 is the common denominator.Convert 2/5 to 4/10:= 4/10 + 3√3/10 = (4 + 3√3)/10So, sin(2α - π/3) = (4 + 3√3)/10Wait, let me check my steps again to make sure I didn't make any mistakes.Starting from sin(2α - π/3) = sin(2α)cos(π/3) - cos(2α)sin(π/3)We have sin(2α) = 4/5, cos(π/3) = 1/2, cos(2α) = -3/5, sin(π/3) = √3/2.So, plugging in:= (4/5)(1/2) - (-3/5)(√3/2)= (4/10) - (-3√3/10)= 4/10 + 3√3/10= (4 + 3√3)/10Yes, that seems correct.So, the value of sin(2α - π/3) is (4 + 3√3)/10.I think that's it. Let me just recap:1. Used the tangent addition formula to find tan α.2. Found sin α and cos α using the right triangle approach since tan α = 2.3. Applied the double-angle formulas to find sin(2α) and cos(2α).4. Used the sine of a difference formula to compute sin(2α - π/3).5. Simplified the expression to get the final result.Everything seems to check out. I don't see any errors in the calculations, so I think these are the correct answers.**Final Answer**(1) boxed{2}(2) boxed{dfrac{4 + 3sqrt{3}}{10}}

💡Alright, so I have this equation: (x - 3) raised to the power of (36 - x²) equals 1. I need to find how many integer solutions exist for this equation. Hmm, okay. Let me think about how to approach this.First, I remember that for an expression a^b = 1, there are a few cases where this can be true. One case is when the exponent b is zero because any non-zero number raised to the power of zero is 1. Another case is when the base a is 1 because 1 raised to any power is still 1. There's also a case when the base a is -1 and the exponent b is even because (-1) raised to an even power is 1. So, I should consider these three scenarios separately.Let me start with the first case: when the exponent is zero. That means 36 - x² = 0. Solving for x, I get x² = 36, which leads to x = 6 or x = -6. Okay, so that gives me two potential solutions: 6 and -6.Next, the second case: when the base is 1. That means x - 3 = 1. Solving for x, I add 3 to both sides and get x = 4. So, x = 4 is another solution.Now, the third case: when the base is -1 and the exponent is even. So, x - 3 = -1, which means x = 2. But I also need to check if the exponent, 36 - x², is even when x = 2. Plugging in x = 2, I get 36 - (2)² = 36 - 4 = 32. 32 is indeed even, so x = 2 is a valid solution.Wait, are there any other cases? Let me think. If the base is any other number, say a, and the exponent is zero, we've already covered that. If the base is 1, we've covered that. If the base is -1 and the exponent is even, we've covered that. Are there any other possibilities where a^b = 1?Hmm, another thought: if the base is not 1 or -1, but the exponent is zero, that's already covered. Also, if the base is a complex number, but since we're dealing with integers, I don't think that applies here. So, I think I've covered all the cases.Let me list out the solutions I've found so far: x = 6, x = -6, x = 4, and x = 2. That's four solutions. But wait, I should double-check each one to make sure they actually satisfy the original equation.Starting with x = 6: Plugging into the equation, (6 - 3)^(36 - 6²) = 3^(36 - 36) = 3^0 = 1. Yep, that works.Next, x = -6: (-6 - 3)^(36 - (-6)²) = (-9)^(36 - 36) = (-9)^0 = 1. That also works.x = 4: (4 - 3)^(36 - 4²) = 1^(36 - 16) = 1^20 = 1. Perfect.x = 2: (2 - 3)^(36 - 2²) = (-1)^(36 - 4) = (-1)^32. Since 32 is even, (-1)^32 = 1. That works too.So, all four solutions are valid. Are there any other integer values of x that I might have missed?Let me think about x = 3. If x = 3, then the base becomes 0, and 0 raised to any positive exponent is 0, which isn't 1. But 0^0 is undefined, so x = 3 isn't a solution.What about x = 5? Let's see: (5 - 3)^(36 - 25) = 2^11 = 2048, which isn't 1. So, x = 5 isn't a solution.x = 1: (1 - 3)^(36 - 1) = (-2)^35. Since 35 is odd, (-2)^35 is negative, not 1. So, x = 1 isn't a solution.x = 0: (0 - 3)^(36 - 0) = (-3)^36. That's a positive number, but it's 3^36, which is definitely not 1. So, x = 0 isn't a solution.x = -1: (-1 - 3)^(36 - 1) = (-4)^35. Again, negative and not 1.x = -2: (-2 - 3)^(36 - 4) = (-5)^32. That's positive, but (-5)^32 is 5^32, which is way larger than 1.x = -3: (-3 - 3)^(36 - 9) = (-6)^27. Negative and not 1.x = -4: (-4 - 3)^(36 - 16) = (-7)^20. Positive, but 7^20 is much larger than 1.x = -5: (-5 - 3)^(36 - 25) = (-8)^11. Negative and not 1.x = -7: (-7 - 3)^(36 - 49) = (-10)^(-13). That's 1/(-10)^13, which is negative and not 1.Wait, x = -7 gives a negative exponent, which would result in a fraction, but since we're dealing with integers, I think negative exponents would result in fractions, which aren't integers, so x = -7 isn't a solution.Similarly, x = 7: (7 - 3)^(36 - 49) = 4^(-13) = 1/(4^13), which is a fraction, not an integer, so x = 7 isn't a solution.So, checking all integers around the ones I found, it seems like only x = 6, -6, 4, and 2 are solutions.Wait, let me think again about the case when the base is -1. I considered x = 2, but is there another x that makes the base -1? Let's see: x - 3 = -1 implies x = 2, which I've already considered. So, no other x would make the base -1.Also, for the exponent being zero, x² = 36, so x = 6 or x = -6, which I have.For the base being 1, x = 4, which I have.Is there any other case where a^b = 1? For example, if a is any root of unity, but since we're dealing with integers, the only roots of unity are 1 and -1. So, I think I've covered all the cases.Therefore, the integer solutions are x = 6, -6, 4, and 2, which makes a total of four solutions.

💡First, I need to determine the probability that William answers at least one of the six questions correctly by guessing. Each question has six possible answers, so the probability of guessing a single question correctly is 1/6. Conversely, the probability of guessing a single question incorrectly is 5/6.Since the questions are independent, the probability of missing all six questions is (5/6) raised to the power of 6.Finally, to find the probability of answering at least one question correctly, I subtract the probability of missing all six questions from 1.

💡Okay, so I have this problem here: Given the function ( f(x) = 3sin x + 2cos x + 1 ). If there are real numbers ( a ), ( b ), ( c ) such that ( af(x) + bf(x - c) = 1 ) holds true for any real number ( x ), then I need to find the value of ( frac{bcos c}{a} ). The options are A: -1, B: 1/2, C: 1, D: -1/2.Hmm, let's try to unpack this step by step. First, I know that ( f(x) ) is a combination of sine and cosine functions with some constants. The equation ( af(x) + bf(x - c) = 1 ) has to hold for any real number ( x ), which means it's an identity. So, the left side must simplify to 1 regardless of the value of ( x ).Let me write down what ( f(x) ) and ( f(x - c) ) are:( f(x) = 3sin x + 2cos x + 1 )( f(x - c) = 3sin(x - c) + 2cos(x - c) + 1 )So, substituting these into the equation ( af(x) + bf(x - c) = 1 ):( a(3sin x + 2cos x + 1) + b(3sin(x - c) + 2cos(x - c) + 1) = 1 )Let me expand this:( 3asin x + 2acos x + a + 3bsin(x - c) + 2bcos(x - c) + b = 1 )Combine like terms:( 3asin x + 2acos x + 3bsin(x - c) + 2bcos(x - c) + (a + b) = 1 )Now, for this equation to hold for all ( x ), the coefficients of the sine and cosine terms must be zero, and the constant term must equal 1. So, let's set up the equations:1. Coefficient of ( sin x ): ( 3a + 3bcos c = 0 )2. Coefficient of ( cos x ): ( 2a - 3bsin c = 0 )3. Coefficient of ( sin(x - c) ): Hmm, wait, actually, I think I might have made a mistake here. Let me think again.Wait, ( sin(x - c) ) and ( cos(x - c) ) can be expanded using angle subtraction formulas:( sin(x - c) = sin x cos c - cos x sin c )( cos(x - c) = cos x cos c + sin x sin c )So, substituting these into the equation:( 3asin x + 2acos x + 3b[sin x cos c - cos x sin c] + 2b[cos x cos c + sin x sin c] + (a + b) = 1 )Now, let's distribute the ( 3b ) and ( 2b ):( 3asin x + 2acos x + 3bsin x cos c - 3bcos x sin c + 2bcos x cos c + 2bsin x sin c + (a + b) = 1 )Now, let's collect like terms for ( sin x ) and ( cos x ):For ( sin x ):( 3a + 3bcos c + 2bsin c )For ( cos x ):( 2a - 3bsin c + 2bcos c )And the constant term:( a + b )So, putting it all together:( [3a + 3bcos c + 2bsin c] sin x + [2a - 3bsin c + 2bcos c] cos x + (a + b) = 1 )Since this equation must hold for all ( x ), the coefficients of ( sin x ) and ( cos x ) must be zero, and the constant term must equal 1. Therefore, we have the following system of equations:1. ( 3a + 3bcos c + 2bsin c = 0 ) (from ( sin x ) coefficient)2. ( 2a - 3bsin c + 2bcos c = 0 ) (from ( cos x ) coefficient)3. ( a + b = 1 ) (from the constant term)So, we have three equations:1. ( 3a + 3bcos c + 2bsin c = 0 )2. ( 2a - 3bsin c + 2bcos c = 0 )3. ( a + b = 1 )Now, let's see if we can solve this system. Maybe we can express ( a ) from equation 3 and substitute into the other equations.From equation 3: ( a = 1 - b )Substitute ( a = 1 - b ) into equations 1 and 2.Equation 1:( 3(1 - b) + 3bcos c + 2bsin c = 0 )Simplify:( 3 - 3b + 3bcos c + 2bsin c = 0 )Combine like terms:( 3 + b(-3 + 3cos c + 2sin c) = 0 )Equation 2:( 2(1 - b) - 3bsin c + 2bcos c = 0 )Simplify:( 2 - 2b - 3bsin c + 2bcos c = 0 )Combine like terms:( 2 + b(-2 - 3sin c + 2cos c) = 0 )So now, we have two equations in terms of ( b ), ( sin c ), and ( cos c ):1. ( 3 + b(-3 + 3cos c + 2sin c) = 0 ) (Equation 1a)2. ( 2 + b(-2 - 3sin c + 2cos c) = 0 ) (Equation 2a)Let me write them as:1. ( b(-3 + 3cos c + 2sin c) = -3 ) (Equation 1b)2. ( b(-2 - 3sin c + 2cos c) = -2 ) (Equation 2b)Let me denote ( S = sin c ) and ( C = cos c ) for simplicity.So, equations become:1. ( b(-3 + 3C + 2S) = -3 ) (Equation 1c)2. ( b(-2 - 3S + 2C) = -2 ) (Equation 2c)Let me write these as:1. ( b(3C + 2S - 3) = -3 ) (Equation 1d)2. ( b(2C - 3S - 2) = -2 ) (Equation 2d)Now, let's denote:Equation 1d: ( b(3C + 2S - 3) = -3 )Equation 2d: ( b(2C - 3S - 2) = -2 )Let me solve for ( b ) from both equations and set them equal.From Equation 1d:( b = frac{-3}{3C + 2S - 3} )From Equation 2d:( b = frac{-2}{2C - 3S - 2} )So, set them equal:( frac{-3}{3C + 2S - 3} = frac{-2}{2C - 3S - 2} )Simplify the negatives:( frac{3}{3C + 2S - 3} = frac{2}{2C - 3S - 2} )Cross-multiplying:( 3(2C - 3S - 2) = 2(3C + 2S - 3) )Expand both sides:Left side: ( 6C - 9S - 6 )Right side: ( 6C + 4S - 6 )Set them equal:( 6C - 9S - 6 = 6C + 4S - 6 )Subtract ( 6C ) and add 6 to both sides:( -9S = 4S )Combine like terms:( -9S - 4S = 0 )( -13S = 0 )So, ( S = 0 ). That is, ( sin c = 0 ).Therefore, ( c ) must be an integer multiple of ( pi ). So, ( c = kpi ) where ( k ) is an integer.Now, since ( sin c = 0 ), ( cos c ) is either 1 or -1.So, ( cos c = pm 1 ).Let's consider both cases.Case 1: ( cos c = 1 )Then, ( c = 2kpi ). Let's substitute ( sin c = 0 ) and ( cos c = 1 ) into our earlier equations.From Equation 1d:( b(3*1 + 2*0 - 3) = b(3 + 0 - 3) = b(0) = -3 )But ( b*0 = -3 ) implies ( 0 = -3 ), which is impossible. So, this case is invalid.Case 2: ( cos c = -1 )Then, ( c = (2k + 1)pi ). Substitute ( sin c = 0 ) and ( cos c = -1 ) into Equation 1d:( b(3*(-1) + 2*0 - 3) = b(-3 + 0 - 3) = b(-6) = -3 )So, ( -6b = -3 ) implies ( b = frac{-3}{-6} = frac{1}{2} ).Similarly, let's check Equation 2d with ( cos c = -1 ) and ( sin c = 0 ):( b(2*(-1) - 3*0 - 2) = b(-2 - 0 - 2) = b(-4) = -2 )So, ( -4b = -2 ) implies ( b = frac{-2}{-4} = frac{1}{2} ).Consistent. So, ( b = frac{1}{2} ).From equation 3: ( a + b = 1 ), so ( a = 1 - b = 1 - frac{1}{2} = frac{1}{2} ).Therefore, ( a = frac{1}{2} ), ( b = frac{1}{2} ), and ( cos c = -1 ).Now, we need to find ( frac{bcos c}{a} ).Substitute the known values:( frac{bcos c}{a} = frac{frac{1}{2} * (-1)}{frac{1}{2}} = frac{-frac{1}{2}}{frac{1}{2}} = -1 ).So, the value is -1, which corresponds to option A.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Expanded ( f(x - c) ) correctly using angle subtraction formulas.2. Collected like terms for ( sin x ) and ( cos x ), set their coefficients to zero, and the constant term to 1.3. Derived three equations, substituted ( a = 1 - b ) into the first two, leading to two equations in ( b ), ( sin c ), and ( cos c ).4. Solved for ( b ) in both equations, set them equal, leading to ( sin c = 0 ).5. Considered both ( cos c = 1 ) and ( cos c = -1 ), found that ( cos c = 1 ) leads to a contradiction, so ( cos c = -1 ).6. Substituted back to find ( b = frac{1}{2} ), then ( a = frac{1}{2} ).7. Calculated ( frac{bcos c}{a} = -1 ).Everything seems to check out. So, the answer should be A: -1.**Final Answer**boxed{A}