answer：Okay, so Wayne wants to serve shrimp cocktail as an appetizer at his event. He’s planning on having 40 guests, and he wants to give each guest 5 shrimp. That sounds like a lot of shrimp, but I guess it’s a nice appetizer. Anyway, I need to figure out how much he’s going to spend on the shrimp.First, I think I should calculate the total number of shrimp he needs. If each guest gets 5 shrimp and there are 40 guests, then it’s just 5 multiplied by 40. Let me do that: 5 times 40 is 200. So, Wayne needs 200 shrimp in total.Now, the next thing is figuring out how many pounds of shrimp that is. The problem says that each pound has 20 shrimp. So, if there are 20 shrimp in a pound, how many pounds does he need to get 200 shrimp? I think I need to divide the total number of shrimp by the number of shrimp per pound. So, 200 divided by 20. Let me calculate that: 200 divided by 20 is 10. So, Wayne needs 10 pounds of shrimp.Okay, now that I know he needs 10 pounds of shrimp, I need to find out how much that will cost him. The shrimp costs 17.00 per pound. So, if one pound is 17, then 10 pounds would be 17 multiplied by 10. Let me do that: 17 times 10 is 170. So, Wayne will spend 170 on the shrimp appetizer.Wait, let me double-check my calculations to make sure I didn’t make any mistakes. Starting with the total number of shrimp: 5 shrimp per guest times 40 guests is indeed 200 shrimp. Then, dividing 200 by 20 shrimp per pound gives me 10 pounds. Multiplying 10 pounds by 17 per pound gives me 170. Yeah, that seems right.I think I’ve got it. Wayne needs to spend 170 on shrimp for his appetizer.

answer：Okay, so I have this geometry problem here about a right-angled sector AOB. Let me try to visualize it first. A right-angled sector means that the angle at O is 90 degrees, right? So, OA and OB are the two radii forming a quarter-circle, I guess.Now, the problem says that an arc OC is drawn from point B as its center, with radius BO. Hmm, so point C is where this new arc intersects the original arc AB. So, arc OC is part of a circle centered at B with radius BO, which is the same as the radius of the sector AOB. That means arc OC is actually a quarter-circle as well, but centered at B instead of O. Interesting.Next, there's a circle ω that's tangent to arc AB, arc OC, and line OA. So, ω is snug against the original quarter-circle AB, the new arc OC, and the line OA. Then, there's another circle ω' that's tangent to arc OC, line OA, and circle ω. I need to find the ratio of the radius of ω to the radius of ω'.Alright, let me start by setting up some coordinates to make this easier. Let's place point O at the origin (0,0). Since it's a right-angled sector, let's assume OA is along the x-axis and OB is along the y-axis. Let's denote the radius of the sector as R. So, point A is at (R, 0) and point B is at (0, R).Now, the arc OC is drawn from point B with radius BO, which is R. So, the center of this arc is at B (0, R), and it has a radius R. The equation of this circle is (x - 0)^2 + (y - R)^2 = R^2. Simplifying, that's x^2 + (y - R)^2 = R^2.The original arc AB is part of the circle centered at O with radius R, so its equation is x^2 + y^2 = R^2. The intersection point C of arcs AB and OC can be found by solving these two equations:1. x^2 + y^2 = R^22. x^2 + (y - R)^2 = R^2Subtracting equation 1 from equation 2:x^2 + (y - R)^2 - (x^2 + y^2) = R^2 - R^2Expanding (y - R)^2: y^2 - 2Ry + R^2So, x^2 + y^2 - 2Ry + R^2 - x^2 - y^2 = 0Simplify: -2Ry + R^2 = 0So, -2Ry + R^2 = 0 => R^2 = 2Ry => y = R/2Plugging y = R/2 back into equation 1: x^2 + (R/2)^2 = R^2 => x^2 + R^2/4 = R^2 => x^2 = (3/4)R^2 => x = ±(√3/2)RSince we're dealing with the sector AOB, which is in the first quadrant, point C must be in the first quadrant as well. So, x is positive, hence x = (√3/2)R. Therefore, point C is at ((√3/2)R, R/2).Alright, so now I know the coordinates of point C. Now, let's think about circle ω. It's tangent to arc AB, arc OC, and line OA. Let me denote the radius of ω as r. The center of ω, let's call it O1, must be somewhere in the first quadrant, near OA.Since ω is tangent to OA, which is the x-axis, the y-coordinate of O1 must be equal to its radius r. So, O1 is at (a, r) for some a > 0.Now, ω is also tangent to arc AB and arc OC. Let's first consider the tangency to arc AB. Arc AB is part of the circle centered at O with radius R. The distance between the centers O and O1 must be equal to R - r because they are tangent internally. So, the distance between (0,0) and (a, r) is √(a^2 + r^2) = R - r.So, we have the equation:√(a^2 + r^2) = R - rSquaring both sides:a^2 + r^2 = (R - r)^2 = R^2 - 2Rr + r^2Simplify:a^2 = R^2 - 2RrSo, a^2 = R(R - 2r)That's one equation.Now, ω is also tangent to arc OC. Arc OC is part of the circle centered at B (0, R) with radius R. The distance between centers O1 (a, r) and B (0, R) must be equal to R + r because they are tangent externally. So, the distance between (a, r) and (0, R) is √(a^2 + (R - r)^2) = R + r.So, we have:√(a^2 + (R - r)^2) = R + rSquaring both sides:a^2 + (R - r)^2 = (R + r)^2Expanding both sides:a^2 + R^2 - 2Rr + r^2 = R^2 + 2Rr + r^2Simplify:a^2 - 2Rr = 2RrSo, a^2 = 4RrBut from earlier, we have a^2 = R(R - 2r). So,R(R - 2r) = 4RrDivide both sides by R (assuming R ≠ 0):R - 2r = 4rSo, R = 6rTherefore, R = 6r. So, the radius of the sector is six times the radius of circle ω.So, r = R/6.Alright, so that's the radius of ω. Now, let's move on to circle ω'. It's tangent to arc OC, line OA, and circle ω. Let's denote the radius of ω' as r'. Its center, let's call it O2, must also be in the first quadrant, near OA.Since ω' is tangent to OA, its y-coordinate must be equal to its radius r'. So, O2 is at (b, r') for some b > 0.Now, ω' is tangent to arc OC, which is centered at B (0, R) with radius R. So, the distance between O2 (b, r') and B (0, R) must be equal to R + r' because they are tangent externally.So, the distance is √(b^2 + (R - r')^2) = R + r'Squaring both sides:b^2 + (R - r')^2 = (R + r')^2Expanding both sides:b^2 + R^2 - 2Rr' + r'^2 = R^2 + 2Rr' + r'^2Simplify:b^2 - 2Rr' = 2Rr'So, b^2 = 4Rr'That's one equation.Additionally, ω' is tangent to circle ω. The distance between their centers O1 (a, r) and O2 (b, r') must be equal to r + r'.So, √((a - b)^2 + (r - r')^2) = r + r'Squaring both sides:(a - b)^2 + (r - r')^2 = (r + r')^2Expanding both sides:(a^2 - 2ab + b^2) + (r^2 - 2rr' + r'^2) = r^2 + 2rr' + r'^2Simplify:a^2 - 2ab + b^2 - 2rr' = 2rr'So, a^2 - 2ab + b^2 = 4rr'But from earlier, we have a^2 = R(R - 2r) and b^2 = 4Rr'Also, since R = 6r, let's substitute R = 6r into these equations.First, a^2 = 6r(6r - 2r) = 6r(4r) = 24r^2So, a = √(24r^2) = 2r√6Similarly, b^2 = 4*6r*r' = 24r r'So, b = √(24r r') = 2√(6r r')Now, plug a and b into the equation a^2 - 2ab + b^2 = 4rr':24r^2 - 2*(2r√6)*(2√(6r r')) + 24r r' = 4rr'Simplify term by term:First term: 24r^2Second term: -2*(2r√6)*(2√(6r r')) = -8r√6 * √(6r r') = -8r√(36r^2 r') = -8r*6r√(r') = -48r^2√(r')Wait, hold on, that seems complicated. Maybe I made a mistake in simplifying.Wait, let's do it step by step.Second term: -2ab = -2*(2r√6)*(2√(6r r')) = -4r√6 * 2√(6r r') = -8r√6 * √(6r r')Wait, no, that's not correct. Let me re-express:ab = (2r√6)*(2√(6r r')) = 4r√6 * √(6r r')So, ab = 4r√6 * √(6r r') = 4r * √(36r^2 r') = 4r * 6r√(r') = 24r^2√(r')Wait, that seems too complicated. Maybe there's a better way.Alternatively, let's express everything in terms of r and r'.We have:a = 2r√6b = 2√(6r r')So, ab = (2r√6)(2√(6r r')) = 4r√6 * √(6r r') = 4r * √(36r^2 r') = 4r * 6r√(r') = 24r^2√(r')So, the second term is -2ab = -48r^2√(r')Third term: b^2 = 24r r'So, putting it all together:24r^2 - 48r^2√(r') + 24r r' = 4r r'Simplify:24r^2 - 48r^2√(r') + 24r r' - 4r r' = 0Simplify terms:24r^2 - 48r^2√(r') + 20r r' = 0Divide both sides by r (assuming r ≠ 0):24r - 48r√(r') + 20r' = 0Let me factor out 4r:4r(6 - 12√(r') + 5r'/r) = 0Wait, that might not be helpful. Alternatively, let's divide the entire equation by r:24 - 48√(r') + 20r'/r = 0Let me denote k = √(r'). Then, r' = k^2, and the equation becomes:24 - 48k + 20k^2/r = 0But since R = 6r, and we might need to express everything in terms of r. Alternatively, let's express r' in terms of r.Wait, maybe it's better to let t = r'/r, so r' = tr. Then, the equation becomes:24r - 48r√(tr) + 20(tr) = 0Divide both sides by r:24 - 48√(t) + 20t = 0So, we have:20t - 48√t + 24 = 0Let me set u = √t, so t = u^2. Then, the equation becomes:20u^2 - 48u + 24 = 0Divide all terms by 4:5u^2 - 12u + 6 = 0Now, solve for u using quadratic formula:u = [12 ± √(144 - 120)] / 10 = [12 ± √24] / 10 = [12 ± 2√6] / 10 = [6 ± √6]/5So, u = (6 + √6)/5 or u = (6 - √6)/5Since u = √t and t = r'/r, which must be positive, both solutions are positive. However, we need to check which one makes sense geometrically.If u = (6 + √6)/5 ≈ (6 + 2.45)/5 ≈ 8.45/5 ≈ 1.69, which would mean r' ≈ 1.69r. But since ω' is tangent to ω and both are inside the sector, r' should be smaller than r. So, this solution might not make sense.Alternatively, u = (6 - √6)/5 ≈ (6 - 2.45)/5 ≈ 3.55/5 ≈ 0.71, which is less than 1, so r' ≈ 0.71r, which is plausible.Therefore, u = (6 - √6)/5, so t = u^2 = [(6 - √6)/5]^2 = (36 - 12√6 + 6)/25 = (42 - 12√6)/25Therefore, r' = t r = [(42 - 12√6)/25] rSo, the ratio r/r' is r / [(42 - 12√6)/25 r] = 25 / (42 - 12√6)To rationalize the denominator, multiply numerator and denominator by (42 + 12√6):25*(42 + 12√6) / [(42 - 12√6)(42 + 12√6)] = [25*(42 + 12√6)] / [42^2 - (12√6)^2]Calculate denominator:42^2 = 1764(12√6)^2 = 144*6 = 864So, denominator = 1764 - 864 = 900So, numerator = 25*(42 + 12√6) = 1050 + 300√6Thus, ratio = (1050 + 300√6)/900 = Simplify by dividing numerator and denominator by 150:(7 + 2√6)/6So, the ratio of the radius of ω to ω' is (7 + 2√6)/6.Wait, but earlier I thought u was (6 - √6)/5, which gave t = [(6 - √6)/5]^2. Let me double-check that calculation.t = [(6 - √6)/5]^2 = (36 - 12√6 + 6)/25 = (42 - 12√6)/25. Yes, that's correct.Then, r' = t r = (42 - 12√6)/25 rSo, r/r' = 25 / (42 - 12√6) = [25*(42 + 12√6)] / [ (42)^2 - (12√6)^2 ] = [25*(42 + 12√6)] / (1764 - 864) = [25*(42 + 12√6)] / 900Simplify numerator: 25*42 = 1050, 25*12√6 = 300√6So, numerator = 1050 + 300√6Divide by 900: (1050 + 300√6)/900 = (7 + 2√6)/6Yes, that's correct.Alternatively, if I consider the other solution u = (6 + √6)/5, then t = [(6 + √6)/5]^2 = (36 + 12√6 + 6)/25 = (42 + 12√6)/25Then, r' = t r = (42 + 12√6)/25 rSo, r/r' = 25 / (42 + 12√6) = [25*(42 - 12√6)] / [ (42)^2 - (12√6)^2 ] = [25*(42 - 12√6)] / 900Which gives (1050 - 300√6)/900 = (7 - 2√6)/6But since (7 - 2√6)/6 is approximately (7 - 4.898)/6 ≈ 2.102/6 ≈ 0.35, which is less than 1, but we already have r' ≈ 0.71r, so this might not make sense.Wait, actually, if r' = (42 + 12√6)/25 r ≈ (42 + 29.39)/25 ≈ 71.39/25 ≈ 2.855r, which is larger than r, which is not possible because ω' is inside ω? Wait, no, ω' is tangent to ω, but not necessarily inside it. It could be outside, but in this configuration, since both are tangent to OA and arc OC, ω' is likely smaller than ω.Wait, perhaps both solutions are valid, one for ω' inside the space between ω and OA, and the other for ω' outside, but in this problem, it's specified that ω' is tangent to arc OC, line OA, and circle ω. So, likely, the smaller solution is the correct one.Therefore, the ratio is (7 + 2√6)/6.Wait, but let me think again. If r' = (42 - 12√6)/25 r ≈ (42 - 29.39)/25 ≈ 12.61/25 ≈ 0.504r, which is about half of r, which makes sense for ω' being smaller.Alternatively, if I use the ratio (7 + 2√6)/6 ≈ (7 + 4.899)/6 ≈ 11.899/6 ≈ 1.983, which is approximately 2, but that would mean r is about twice r', which is consistent with r' being smaller.Wait, no, the ratio r/r' is (7 + 2√6)/6 ≈ 1.983, so r ≈ 1.983 r', meaning r' ≈ r / 1.983 ≈ 0.504r, which matches the earlier approximation.So, yes, the ratio is (7 + 2√6)/6.But wait, the problem asks for the ratio of the radius of ω to ω', so r/r' = (7 + 2√6)/6.Alternatively, sometimes problems expect the ratio in the form of a fraction with integers, but in this case, it's already simplified.So, the final answer is (7 + 2√6)/6.But let me check if I made any mistakes in the algebra.When I had 24r - 48r√(r') + 20r r' = 0, I divided by r to get 24 - 48√(r') + 20r'/r = 0.Then, set t = r'/r, so 24 - 48√t + 20t = 0.Then, set u = √t, so 20u^2 - 48u + 24 = 0.Divide by 4: 5u^2 - 12u + 6 = 0.Solution: u = [12 ± √(144 - 120)] / 10 = [12 ± √24]/10 = [12 ± 2√6]/10 = [6 ± √6]/5.Yes, that's correct.So, u = (6 ± √6)/5, so t = u^2 = [(6 ± √6)/5]^2.So, t = (36 ± 12√6 + 6)/25 = (42 ± 12√6)/25.Thus, r' = t r = (42 ± 12√6)/25 r.Therefore, r/r' = 25 / (42 ± 12√6).We chose the negative sign because it gives a smaller r', so r/r' = 25 / (42 - 12√6).Rationalizing:25*(42 + 12√6) / (42^2 - (12√6)^2) = 25*(42 + 12√6)/900 = (1050 + 300√6)/900 = (7 + 2√6)/6.Yes, that's correct.So, the ratio is (7 + 2√6)/6.Alternatively, sometimes problems might expect the answer in a different form, but this seems to be the simplest form.Therefore, the ratio of the radius of circle ω to the radius of circle ω' is (7 + 2√6)/6.

answer：Okay, so I have this problem here: I need to find the greatest integer value of ( b ) so that 5 is not in the range of the function ( y = x^2 + bx + 20 ). Hmm, let me think about how to approach this.First, I remember that the range of a quadratic function ( y = ax^2 + bx + c ) depends on whether it opens upwards or downwards. Since the coefficient of ( x^2 ) here is 1, which is positive, the parabola opens upwards. That means the function has a minimum point, and the range is from that minimum value to infinity. So, if 5 is not in the range, it must be below this minimum value.Wait, actually, no. If the parabola opens upwards, the range is from the minimum value upwards. So, if 5 is not in the range, it must be less than the minimum value. But wait, that doesn't make sense because if the minimum is, say, 6, then the range would be from 6 to infinity, so 5 wouldn't be in the range. But if the minimum is 4, then the range would include 5. So, I need to ensure that the minimum value of the function is greater than 5.Alternatively, another way to think about it is that if 5 is not in the range, then the equation ( x^2 + bx + 20 = 5 ) has no real solutions. That makes sense because if there are no real solutions, it means the function never actually reaches 5. So, let me set up that equation:( x^2 + bx + 20 = 5 )Subtracting 5 from both sides gives:( x^2 + bx + 15 = 0 )Now, for this quadratic equation to have no real solutions, the discriminant must be negative. The discriminant ( D ) of a quadratic equation ( ax^2 + bx + c = 0 ) is given by ( D = b^2 - 4ac ). If ( D < 0 ), there are no real solutions.In this case, ( a = 1 ), ( b = b ), and ( c = 15 ). So, plugging into the discriminant formula:( D = b^2 - 4(1)(15) = b^2 - 60 )We need this discriminant to be less than zero:( b^2 - 60 < 0 )Solving for ( b ):( b^2 < 60 )Taking the square root of both sides:( |b| < sqrt{60} )Calculating ( sqrt{60} ):( sqrt{60} approx 7.746 )So, ( b ) must be greater than ( -7.746 ) and less than ( 7.746 ). But the question asks for the greatest integer value of ( b ). Since ( 7.746 ) is approximately 7.746, the greatest integer less than this is 7.Wait a second, let me double-check. If ( b = 7 ), then the discriminant would be ( 7^2 - 60 = 49 - 60 = -11 ), which is negative, so no real solutions. That means 5 is not in the range. If ( b = 8 ), then the discriminant would be ( 8^2 - 60 = 64 - 60 = 4 ), which is positive, meaning there are real solutions, so 5 would be in the range. Therefore, 8 is too big, and 7 is the largest integer that works.So, yeah, I think 7 is the correct answer. Let me just recap to make sure I didn't miss anything.1. Start with the function ( y = x^2 + bx + 20 ).2. To ensure 5 is not in the range, set ( x^2 + bx + 20 = 5 ) and ensure no real solutions.3. Simplify to ( x^2 + bx + 15 = 0 ).4. Calculate discriminant ( D = b^2 - 60 ).5. For no real solutions, ( D < 0 ) so ( b^2 < 60 ).6. ( b ) must be between ( -sqrt{60} ) and ( sqrt{60} ), approximately between -7.746 and 7.746.7. The greatest integer less than 7.746 is 7.8. Verify by plugging ( b = 7 ) and ( b = 8 ) into the discriminant to ensure the result.Everything seems to check out. I don't think I made any mistakes in the reasoning. So, I'm confident that 7 is the correct answer.**Final Answer**The greatest integer value of ( b ) is boxed{7}.

answer：Okay, so I have this problem where the sum of two numbers, x and y, is 399, and the fraction x over y is 0.9. I need to find the value of y minus x. Hmm, let's see how to approach this.First, I know that x plus y equals 399. That's straightforward. And then, x divided by y is 0.9. So, maybe I can express x in terms of y using the second equation and then substitute it into the first equation to find the values of x and y.Alright, so from the second equation, x over y is 0.9. That means x equals 0.9 times y. So, x = 0.9y. Got that.Now, I can plug this expression for x into the first equation. So, substituting x with 0.9y in the first equation gives me 0.9y plus y equals 399. Let me write that down: 0.9y + y = 399.Combining like terms, 0.9y plus y is 1.9y. So, 1.9y equals 399. To find y, I need to divide both sides by 1.9. Let me do that: y equals 399 divided by 1.9.Hmm, calculating that. Let me see, 399 divided by 1.9. Maybe I can simplify this division. If I multiply both numerator and denominator by 10 to eliminate the decimal, that would give me 3990 divided by 19. Let me try that.So, 3990 divided by 19. Let's see, 19 times 200 is 3800. Subtracting that from 3990 gives me 190. Then, 19 times 10 is 190. So, adding that to 200 gives me 210. So, y equals 210.Now that I have y, I can find x using the expression x = 0.9y. Plugging in y = 210, x equals 0.9 times 210. Let me calculate that: 0.9 times 200 is 180, and 0.9 times 10 is 9, so adding those together gives me 189. So, x is 189.Finally, I need to find y minus x. So, y is 210 and x is 189. Subtracting 189 from 210 gives me 21. Therefore, y minus x is 21.Wait, let me double-check my calculations to make sure I didn't make any mistakes. Starting from the beginning: x + y = 399, and x/y = 0.9. So, x = 0.9y. Substituting into the first equation: 0.9y + y = 1.9y = 399. So, y = 399 / 1.9 = 210. Then, x = 0.9 * 210 = 189. And y - x = 210 - 189 = 21. Yeah, that seems correct.I think I've got it right. The value of y minus x is 21.

answer：Alright, I have this equation to solve: (frac{|cos x| - cos 3x}{cos x sin 2x} = frac{2}{sqrt{3}}). Hmm, it looks a bit complicated, but I think I can break it down step by step.First, I notice that there's an absolute value around (cos x), which means I might have to consider two cases: one where (cos x) is non-negative and another where it's negative. That makes sense because the absolute value will change depending on the sign of (cos x).Let me write down the equation again:[frac{|cos x| - cos 3x}{cos x sin 2x} = frac{2}{sqrt{3}}]Okay, so I'll start by considering the two cases for (cos x).**Case 1: (cos x geq 0)**In this case, (|cos x| = cos x), so the equation simplifies to:[frac{cos x - cos 3x}{cos x sin 2x} = frac{2}{sqrt{3}}]I remember that (cos 3x) can be expressed using a trigonometric identity. Let me recall it: (cos 3x = 4cos^3 x - 3cos x). Yeah, that's right. So substituting that into the equation:[frac{cos x - (4cos^3 x - 3cos x)}{cos x sin 2x} = frac{2}{sqrt{3}}]Simplifying the numerator:[cos x - 4cos^3 x + 3cos x = 4cos x - 4cos^3 x]So now the equation becomes:[frac{4cos x - 4cos^3 x}{cos x sin 2x} = frac{2}{sqrt{3}}]I can factor out (4cos x) from the numerator:[frac{4cos x (1 - cos^2 x)}{cos x sin 2x} = frac{2}{sqrt{3}}]Oh, (1 - cos^2 x) is (sin^2 x), so:[frac{4cos x sin^2 x}{cos x sin 2x} = frac{2}{sqrt{3}}]The (cos x) terms cancel out:[frac{4sin^2 x}{sin 2x} = frac{2}{sqrt{3}}]I also remember that (sin 2x = 2sin x cos x), so substituting that in:[frac{4sin^2 x}{2sin x cos x} = frac{2}{sqrt{3}}]Simplifying the fraction:[frac{2sin x}{cos x} = frac{2}{sqrt{3}}]Which simplifies to:[tan x = frac{1}{sqrt{3}}]So, (x = frac{pi}{6} + kpi), where (k) is any integer. But wait, I need to remember that in this case, (cos x geq 0). So, (cos x) is non-negative in the first and fourth quadrants. Therefore, the solutions here would be (x = frac{pi}{6} + 2kpi) and (x = frac{7pi}{6} + 2kpi). Hmm, but (frac{7pi}{6}) is in the third quadrant where (cos x) is negative, so actually, only (x = frac{pi}{6} + 2kpi) is valid here.Wait, no. Let me think again. If (x = frac{pi}{6} + kpi), then when (k) is even, it's in the first quadrant, and when (k) is odd, it's in the third quadrant where (cos x) is negative. So, actually, only the solutions where (k) is even are valid in this case. So, (x = frac{pi}{6} + 2kpi).Got it. So, that's the solution for Case 1.**Case 2: (cos x < 0)**In this case, (|cos x| = -cos x), so the original equation becomes:[frac{-cos x - cos 3x}{cos x sin 2x} = frac{2}{sqrt{3}}]Again, using the identity (cos 3x = 4cos^3 x - 3cos x), substitute that in:[frac{-cos x - (4cos^3 x - 3cos x)}{cos x sin 2x} = frac{2}{sqrt{3}}]Simplify the numerator:[-cos x - 4cos^3 x + 3cos x = -4cos x - 4cos^3 x]So the equation becomes:[frac{-4cos x - 4cos^3 x}{cos x sin 2x} = frac{2}{sqrt{3}}]Factor out (-4cos x) from the numerator:[frac{-4cos x (1 + cos^2 x)}{cos x sin 2x} = frac{2}{sqrt{3}}]The (cos x) terms cancel out:[frac{-4(1 + cos^2 x)}{sin 2x} = frac{2}{sqrt{3}}]Hmm, this seems a bit more complicated. Let me see. I know that (1 + cos^2 x) doesn't simplify easily, but maybe I can express everything in terms of sine or cosine.Wait, I also remember that (sin 2x = 2sin x cos x). Let me substitute that in:[frac{-4(1 + cos^2 x)}{2sin x cos x} = frac{2}{sqrt{3}}]Simplify the fraction:[frac{-2(1 + cos^2 x)}{sin x cos x} = frac{2}{sqrt{3}}]Divide both sides by 2:[frac{-(1 + cos^2 x)}{sin x cos x} = frac{1}{sqrt{3}}]Hmm, this is getting tricky. Maybe I can express (1 + cos^2 x) in terms of (sin^2 x). Since (1 = sin^2 x + cos^2 x), so:[1 + cos^2 x = sin^2 x + 2cos^2 x]Not sure if that helps. Alternatively, maybe I can write everything in terms of (tan x). Let me try that.Let me denote (t = tan x). Then, (sin x = frac{t}{sqrt{1 + t^2}}) and (cos x = frac{1}{sqrt{1 + t^2}}).Substituting these into the equation:[frac{-(1 + frac{1}{1 + t^2})}{frac{t}{sqrt{1 + t^2}} cdot frac{1}{sqrt{1 + t^2}}} = frac{1}{sqrt{3}}]Simplify the numerator:[1 + frac{1}{1 + t^2} = frac{(1 + t^2) + 1}{1 + t^2} = frac{2 + t^2}{1 + t^2}]So the numerator becomes (-frac{2 + t^2}{1 + t^2}).The denominator:[frac{t}{sqrt{1 + t^2}} cdot frac{1}{sqrt{1 + t^2}} = frac{t}{1 + t^2}]So the entire fraction is:[frac{-frac{2 + t^2}{1 + t^2}}{frac{t}{1 + t^2}} = frac{-(2 + t^2)}{t} = frac{1}{sqrt{3}}]So:[frac{-(2 + t^2)}{t} = frac{1}{sqrt{3}}]Multiply both sides by (t):[-(2 + t^2) = frac{t}{sqrt{3}}]Multiply both sides by (-1):[2 + t^2 = -frac{t}{sqrt{3}}]Bring all terms to one side:[t^2 + frac{t}{sqrt{3}} + 2 = 0]This is a quadratic equation in terms of (t). Let me write it as:[t^2 + frac{1}{sqrt{3}} t + 2 = 0]To solve for (t), use the quadratic formula:[t = frac{-frac{1}{sqrt{3}} pm sqrt{left(frac{1}{sqrt{3}}right)^2 - 4 cdot 1 cdot 2}}{2 cdot 1}]Calculate the discriminant:[left(frac{1}{sqrt{3}}right)^2 - 8 = frac{1}{3} - 8 = -frac{23}{3}]Since the discriminant is negative, there are no real solutions for (t). That means in this case, there are no solutions where (cos x < 0).Wait, that's interesting. So, in Case 2, there are no real solutions because the quadratic equation has no real roots. Therefore, all solutions must come from Case 1.But hold on, let me double-check my steps in Case 2 to make sure I didn't make a mistake.Starting from:[frac{-(1 + cos^2 x)}{sin x cos x} = frac{1}{sqrt{3}}]I tried substituting (t = tan x), which led to a quadratic with no real solutions. Maybe there's another way to approach this.Alternatively, let's express everything in terms of (sin x) and (cos x):[frac{-(1 + cos^2 x)}{sin x cos x} = frac{1}{sqrt{3}}]Multiply both sides by (sin x cos x):[-(1 + cos^2 x) = frac{sin x cos x}{sqrt{3}}]Hmm, not sure if that helps. Maybe square both sides to eliminate the square roots, but that might complicate things further.Alternatively, let's consider that (cos x < 0), so (x) is in the second or third quadrant. Maybe I can express (x) in terms of reference angles.But honestly, since the quadratic in (t) had no real solutions, it seems like there are no solutions in this case. So, all solutions must come from Case 1.But wait, in Case 1, I only found (x = frac{pi}{6} + 2kpi). But I feel like there might be more solutions. Let me think again.When I solved for (tan x = frac{1}{sqrt{3}}), the general solution is (x = frac{pi}{6} + kpi). But since (cos x geq 0), we need to consider where (cos x) is non-negative, which is in the first and fourth quadrants.So, (x = frac{pi}{6} + 2kpi) is in the first quadrant, and (x = frac{7pi}{6} + 2kpi) is in the third quadrant where (cos x) is negative. So, actually, only (x = frac{pi}{6} + 2kpi) is valid in this case.But wait, (frac{7pi}{6}) is in the third quadrant, but (cos x) is negative there, so that solution doesn't belong to Case 1. So, yeah, only (x = frac{pi}{6} + 2kpi) is valid.But I have a feeling I might be missing some solutions. Let me check the original equation again.Original equation:[frac{|cos x| - cos 3x}{cos x sin 2x} = frac{2}{sqrt{3}}]If I plug in (x = frac{pi}{6}):[frac{|cos frac{pi}{6}| - cos frac{pi}{2}}{cos frac{pi}{6} sin frac{pi}{3}} = frac{frac{sqrt{3}}{2} - 0}{frac{sqrt{3}}{2} cdot frac{sqrt{3}}{2}} = frac{frac{sqrt{3}}{2}}{frac{3}{4}} = frac{sqrt{3}}{2} cdot frac{4}{3} = frac{2sqrt{3}}{3} = frac{2}{sqrt{3}}]Yes, that works. What about (x = frac{5pi}{6})?[frac{|cos frac{5pi}{6}| - cos frac{5pi}{2}}{cos frac{5pi}{6} sin frac{5pi}{3}} = frac{frac{sqrt{3}}{2} - 0}{-frac{sqrt{3}}{2} cdot (-frac{sqrt{3}}{2})} = frac{frac{sqrt{3}}{2}}{frac{3}{4}} = frac{2sqrt{3}}{3} = frac{2}{sqrt{3}}]Wait, that also works. But in this case, (cos x = -frac{sqrt{3}}{2}), which is negative, so it should belong to Case 2. But in Case 2, we found no solutions. Hmm, that's confusing.So, (x = frac{5pi}{6}) is a solution, but according to my earlier reasoning, it shouldn't be. Maybe I made a mistake in Case 2.Let me go back to Case 2.**Re-examining Case 2: (cos x < 0)**Starting from:[frac{-cos x - cos 3x}{cos x sin 2x} = frac{2}{sqrt{3}}]Using (cos 3x = 4cos^3 x - 3cos x):[frac{-cos x - (4cos^3 x - 3cos x)}{cos x sin 2x} = frac{2}{sqrt{3}}]Simplify numerator:[-cos x - 4cos^3 x + 3cos x = -4cos^3 x + 2cos x]So the equation becomes:[frac{-4cos^3 x + 2cos x}{cos x sin 2x} = frac{2}{sqrt{3}}]Factor out (-2cos x) from the numerator:[frac{-2cos x (2cos^2 x - 1)}{cos x sin 2x} = frac{2}{sqrt{3}}]Cancel (cos x):[frac{-2(2cos^2 x - 1)}{sin 2x} = frac{2}{sqrt{3}}]Simplify the numerator:[-2(2cos^2 x - 1) = -4cos^2 x + 2]So:[frac{-4cos^2 x + 2}{sin 2x} = frac{2}{sqrt{3}}]Hmm, maybe I can express (2cos^2 x) in terms of (cos 2x). Recall that (2cos^2 x = 1 + cos 2x). So:[-4cos^2 x + 2 = -2(2cos^2 x) + 2 = -2(1 + cos 2x) + 2 = -2 - 2cos 2x + 2 = -2cos 2x]So the equation becomes:[frac{-2cos 2x}{sin 2x} = frac{2}{sqrt{3}}]Simplify:[-2cot 2x = frac{2}{sqrt{3}}]Divide both sides by 2:[-cot 2x = frac{1}{sqrt{3}}]Which implies:[cot 2x = -frac{1}{sqrt{3}}]So, (tan 2x = -sqrt{3}).Therefore, (2x = frac{2pi}{3} + kpi), so (x = frac{pi}{3} + frac{kpi}{2}).Now, considering (cos x < 0), let's find the values of (x) where (cos x) is negative.The general solution is (x = frac{pi}{3} + frac{kpi}{2}).Let's list some solutions:- For (k = 0): (x = frac{pi}{3}). (cos frac{pi}{3} = frac{1}{2} > 0). Not valid.- For (k = 1): (x = frac{pi}{3} + frac{pi}{2} = frac{5pi}{6}). (cos frac{5pi}{6} = -frac{sqrt{3}}{2} < 0). Valid.- For (k = 2): (x = frac{pi}{3} + pi = frac{4pi}{3}). (cos frac{4pi}{3} = -frac{1}{2} < 0). Valid.- For (k = 3): (x = frac{pi}{3} + frac{3pi}{2} = frac{11pi}{6}). (cos frac{11pi}{6} = frac{sqrt{3}}{2} > 0). Not valid.So, the valid solutions in this case are (x = frac{5pi}{6} + 2kpi) and (x = frac{4pi}{3} + 2kpi).Wait, but when (k = 1), (x = frac{5pi}{6}), and when (k = 2), (x = frac{4pi}{3}). So, the general solution is (x = frac{5pi}{6} + 2kpi) and (x = frac{4pi}{3} + 2kpi).But let me check (x = frac{4pi}{3}):[frac{|cos frac{4pi}{3}| - cos 4pi}{cos frac{4pi}{3} sin frac{8pi}{3}} = frac{frac{1}{2} - 1}{-frac{1}{2} cdot frac{sqrt{3}}{2}} = frac{-frac{1}{2}}{-frac{sqrt{3}}{4}} = frac{-frac{1}{2}}{-frac{sqrt{3}}{4}} = frac{2}{sqrt{3}}]Yes, that works. So, (x = frac{4pi}{3}) is also a solution.Therefore, in Case 2, we have solutions at (x = frac{5pi}{6} + 2kpi) and (x = frac{4pi}{3} + 2kpi).So, combining both cases, the solutions are:- From Case 1: (x = frac{pi}{6} + 2kpi)- From Case 2: (x = frac{5pi}{6} + 2kpi) and (x = frac{4pi}{3} + 2kpi)Wait, but (x = frac{4pi}{3}) is actually in the third quadrant where (cos x) is negative, so it's covered in Case 2. Similarly, (x = frac{5pi}{6}) is in the second quadrant where (cos x) is negative, so also covered in Case 2.But earlier, in Case 1, I only found (x = frac{pi}{6} + 2kpi). So, altogether, the solutions are:[x = frac{pi}{6} + 2kpi, quad x = frac{5pi}{6} + 2kpi, quad x = frac{4pi}{3} + 2kpi, quad k in mathbb{Z}]Let me verify (x = frac{5pi}{6}):[frac{|cos frac{5pi}{6}| - cos frac{15pi}{6}}{cos frac{5pi}{6} sin frac{10pi}{6}} = frac{frac{sqrt{3}}{2} - cos frac{5pi}{2}}{-frac{sqrt{3}}{2} cdot sin frac{5pi}{3}} = frac{frac{sqrt{3}}{2} - 0}{-frac{sqrt{3}}{2} cdot (-frac{sqrt{3}}{2})} = frac{frac{sqrt{3}}{2}}{frac{3}{4}} = frac{2sqrt{3}}{3} = frac{2}{sqrt{3}}]Yes, that works.And (x = frac{4pi}{3}):[frac{|cos frac{4pi}{3}| - cos 4pi}{cos frac{4pi}{3} sin frac{8pi}{3}} = frac{frac{1}{2} - 1}{-frac{1}{2} cdot frac{sqrt{3}}{2}} = frac{-frac{1}{2}}{-frac{sqrt{3}}{4}} = frac{2}{sqrt{3}}]That also works.So, putting it all together, the solutions are:[x = frac{pi}{6} + 2kpi, quad x = frac{5pi}{6} + 2kpi, quad x = frac{4pi}{3} + 2kpi, quad k in mathbb{Z}]I think that's all the solutions. Let me just make sure I didn't miss any other quadrants or cases.In Case 1, (cos x geq 0), so solutions are in the first and fourth quadrants. But when I solved, I only got (x = frac{pi}{6} + 2kpi). The other solution (x = frac{7pi}{6} + 2kpi) is in the third quadrant where (cos x) is negative, so it's not included in Case 1.In Case 2, (cos x < 0), so solutions are in the second and third quadrants, which gave me (x = frac{5pi}{6} + 2kpi) and (x = frac{4pi}{3} + 2kpi).Therefore, combining all valid solutions, we have the three sets above.I think that's comprehensive. I don't see any other possible solutions, and I've verified each one by plugging it back into the original equation. So, I'm confident these are all the solutions.

answer：First, I need to determine the percentage of sunflower in Brand A. I know that Brand A is 40% millet and some percentage of sunflower. Since the total percentage for Brand A must add up to 100%, I can set up the equation 40% + Sunflower% = 100%.Next, I'll solve for Sunflower% by subtracting 40% from both sides of the equation:Sunflower% = 100% - 40% = 60%.Therefore, Brand A contains 60% sunflower.