answer：Okay, so I have this problem with two parts. The first part is to prove an inequality involving some constants and variables, and the second part is to use that inequality to find the minimum value of a function. Let me try to work through each part step by step.Starting with part (1): I need to prove that for positive constants a and b (with a not equal to b), and positive variables x and y, the following inequality holds:frac{a^2}{x} + frac{b^2}{y} geq frac{(a + b)^2}{x + y}And also state when equality occurs.Hmm, this looks familiar. I think it might be related to the Cauchy-Schwarz inequality or maybe the AM-GM inequality. Let me recall both.The Cauchy-Schwarz inequality in its basic form is:(u_1v_1 + u_2v_2)^2 leq (u_1^2 + u_2^2)(v_1^2 + v_2^2)But I'm not sure if that's directly applicable here. Maybe if I set up the terms appropriately.Alternatively, the AM-GM inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. But in this case, we have fractions with variables in the denominators, so maybe I need to manipulate the terms to apply AM-GM.Wait, another thought: there's an inequality called the Titu's lemma, which is a specific case of the Cauchy-Schwarz inequality. It states that:frac{a_1^2}{b_1} + frac{a_2^2}{b_2} + dots + frac{a_n^2}{b_n} geq frac{(a_1 + a_2 + dots + a_n)^2}{b_1 + b_2 + dots + b_n}for positive real numbers a_i and b_i. That seems exactly like what we have here. So in our case, n=2, a1=a, a2=b, b1=x, b2=y.Therefore, applying Titu's lemma directly gives:frac{a^2}{x} + frac{b^2}{y} geq frac{(a + b)^2}{x + y}So that proves the inequality. Now, for the equality condition. In Titu's lemma, equality holds if and only if the ratios of the corresponding terms are equal. That is,frac{a}{x} = frac{b}{y}So, equality holds when a/x equals b/y.Alternatively, if I think about the Cauchy-Schwarz inequality, equality holds when the vectors are proportional. So, in this case, the vectors (a, b) and (1/x, 1/y) must be proportional, which again leads to a/x = b/y.So, that's part (1) done.Moving on to part (2): Using the conclusion from part (1), I need to find the minimum value of the functionf(x) = frac{2}{x} + frac{9}{1 - 2x}where x is in the interval (0, 1/2). Also, I need to specify the value of x where this minimum occurs.Alright, let's see. The function f(x) is given as 2/x + 9/(1 - 2x). I notice that this resembles the left-hand side of the inequality we proved in part (1). Let me try to match the terms.In part (1), we had:frac{a^2}{x} + frac{b^2}{y} geq frac{(a + b)^2}{x + y}Comparing this with f(x), I can think of 2/x as (sqrt(2))^2 / x and 9/(1 - 2x) as (3)^2 / (1 - 2x). So, if I set a = sqrt(2) and b = 3, then:frac{(sqrt{2})^2}{x} + frac{3^2}{y} = frac{2}{x} + frac{9}{y}But in our function, the second term is 9/(1 - 2x), so y would be 1 - 2x. So, let me set y = 1 - 2x.Therefore, applying the inequality from part (1):frac{2}{x} + frac{9}{1 - 2x} geq frac{(sqrt{2} + 3)^2}{x + (1 - 2x)}Simplify the denominator on the right-hand side:x + (1 - 2x) = 1 - xSo, the inequality becomes:frac{2}{x} + frac{9}{1 - 2x} geq frac{(sqrt{2} + 3)^2}{1 - x}Wait, but that doesn't seem to directly give me a numerical minimum. Maybe I made a wrong substitution. Let me think again.Alternatively, perhaps I should set up the inequality so that the denominators add up to something specific. Let me see.In part (1), the denominators were x and y, which added up to x + y. In our function, the denominators are x and 1 - 2x, which add up to 1 - x. Hmm, but in the inequality, the right-hand side is (a + b)^2 over x + y.Wait, maybe I can set y = 1 - 2x, so that x + y = 1 - x, as I did before. Then, the right-hand side is (sqrt(2) + 3)^2 / (1 - x). But I don't know what 1 - x is in terms of the function. Maybe this approach isn't the best.Wait another thought: perhaps I can adjust the variables so that the denominators add up to 1. Let me try to manipulate the function.Let me denote u = x and v = 1 - 2x. Then, u + v = 1 - x. Hmm, but that still leaves me with 1 - x, which is not a constant.Alternatively, maybe I can write the function in terms of two variables that add up to a constant. Let me think.Suppose I let y = 1 - 2x, so that x + y = 1 - x. But as before, that doesn't seem helpful.Wait, perhaps I can adjust the coefficients so that the denominators can be expressed as multiples of some variable. Let me try to see.Let me consider the function f(x) = 2/x + 9/(1 - 2x). Let me denote t = x, so that 1 - 2x = 1 - 2t. Then, f(t) = 2/t + 9/(1 - 2t). Hmm, not sure if that helps.Wait, another idea: perhaps I can use substitution to make the denominators add up to 1. Let me set u = x and v = (1 - 2x)/2. Then, u + v = x + (1 - 2x)/2 = (2x + 1 - 2x)/2 = 1/2. So, u + v = 1/2.But then, f(x) = 2/u + 9/(2v). Hmm, let's see:f(x) = 2/u + 9/(2v) = 2/u + (9/2)/v.But then, if I set a = sqrt(2) and b = sqrt(9/2) = 3/sqrt(2), then:( a^2 ) / u + ( b^2 ) / v = 2/u + (9/2)/v.So, applying the inequality from part (1):2/u + (9/2)/v ≥ (sqrt(2) + 3/sqrt(2))^2 / (u + v) = (sqrt(2) + 3/sqrt(2))^2 / (1/2)Let me compute that:First, sqrt(2) + 3/sqrt(2) = sqrt(2) + (3 sqrt(2))/2 = (2 sqrt(2) + 3 sqrt(2))/2 = (5 sqrt(2))/2So, (sqrt(2) + 3/sqrt(2))^2 = (5 sqrt(2)/2)^2 = (25 * 2)/4 = 50/4 = 12.5Therefore, the right-hand side is 12.5 / (1/2) = 25.So, f(x) = 2/u + 9/(2v) ≥ 25.Therefore, the minimum value of f(x) is 25.But wait, let me check if this substitution is correct.I set u = x and v = (1 - 2x)/2, so that u + v = 1/2. Then, f(x) = 2/u + 9/(2v). So, I can write this as (sqrt(2))^2 / u + (3/sqrt(2))^2 / v, which fits the form of part (1). Therefore, applying the inequality, f(x) ≥ (sqrt(2) + 3/sqrt(2))^2 / (u + v) = (5 sqrt(2)/2)^2 / (1/2) = 25.So, the minimum value is 25.Now, to find the value of x where this minimum occurs, we need to satisfy the equality condition from part (1), which is a/u = b/v.In our substitution, a = sqrt(2), b = 3/sqrt(2), u = x, v = (1 - 2x)/2.So, equality holds when:sqrt(2)/x = (3/sqrt(2)) / [(1 - 2x)/2]Simplify the right-hand side:(3/sqrt(2)) / [(1 - 2x)/2] = (3/sqrt(2)) * (2)/(1 - 2x) = (6/sqrt(2))/(1 - 2x) = (3 sqrt(2))/(1 - 2x)So, we have:sqrt(2)/x = 3 sqrt(2)/(1 - 2x)We can cancel sqrt(2) from both sides:1/x = 3/(1 - 2x)Cross-multiplying:1 - 2x = 3x1 = 5xx = 1/5So, the minimum occurs at x = 1/5.Let me verify this result.If x = 1/5, then 1 - 2x = 1 - 2*(1/5) = 1 - 2/5 = 3/5.So, f(1/5) = 2/(1/5) + 9/(3/5) = 10 + 15 = 25, which matches our earlier conclusion.Therefore, the minimum value of f(x) is 25, achieved at x = 1/5.I think that's it. I used the inequality from part (1) by appropriately substituting variables to fit the form, applied the inequality, and then found the condition for equality to determine the value of x where the minimum occurs.

answer：Okay, so I need to calculate the value of this expression: sin(4π/3) multiplied by cos(5π/6) multiplied by tan(-4π/3). Hmm, that seems a bit complicated at first glance, but I think I can break it down step by step.First, let me recall some basic trigonometric identities and properties. I remember that sine and cosine functions have specific values at certain standard angles, and tangent is just sine over cosine. Also, I should remember the signs of these functions in different quadrants because the angles given here are not in the first quadrant.Let me start by analyzing each trigonometric function separately.Starting with sin(4π/3). I know that 4π/3 is more than π (which is 3π/3), so it's in the third quadrant. In the third quadrant, sine is negative. To find the reference angle, I subtract π from 4π/3, which gives me π/3. So, sin(4π/3) is equal to -sin(π/3). I remember that sin(π/3) is √3/2, so sin(4π/3) should be -√3/2.Next, cos(5π/6). Similarly, 5π/6 is in the second quadrant because it's between π/2 and π. In the second quadrant, cosine is negative. The reference angle here is π - 5π/6, which is π/6. So, cos(5π/6) is equal to -cos(π/6). I recall that cos(π/6) is √3/2, so cos(5π/6) is -√3/2.Now, tan(-4π/3). Tangent is a periodic function with a period of π, so tan(-4π/3) is the same as tan(-4π/3 + 2π) which simplifies to tan(2π/3). Alternatively, I can think of it as tan(-4π/3) = -tan(4π/3) because tangent is an odd function. Let me go with the second approach because it might be simpler.So, tan(4π/3) is in the third quadrant where tangent is positive. The reference angle is again π/3, so tan(4π/3) is tan(π/3) which is √3. Therefore, tan(-4π/3) is -√3.Now, putting it all together:sin(4π/3) = -√3/2cos(5π/6) = -√3/2tan(-4π/3) = -√3So, multiplying them together:(-√3/2) * (-√3/2) * (-√3)Let me compute this step by step.First, multiply the first two terms:(-√3/2) * (-√3/2) = (√3 * √3)/(2 * 2) = (3)/4But since both terms are negative, their product is positive. So, that gives me 3/4.Now, multiply this result by the third term, which is -√3:(3/4) * (-√3) = -3√3/4So, the final result is -3√3/4.Let me double-check my steps to make sure I didn't make any mistakes.1. sin(4π/3): Third quadrant, reference angle π/3, sine is negative. So, -√3/2. Correct.2. cos(5π/6): Second quadrant, reference angle π/6, cosine is negative. So, -√3/2. Correct.3. tan(-4π/3): Using tan(-x) = -tan(x), so -tan(4π/3). tan(4π/3) is in the third quadrant, reference angle π/3, tangent is positive. So, tan(4π/3) = √3, hence tan(-4π/3) = -√3. Correct.Multiplying them:(-√3/2) * (-√3/2) = (√3 * √3)/(2 * 2) = 3/4. Then, 3/4 * (-√3) = -3√3/4. That seems right.I think I did everything correctly. So, the value of the expression is -3√3/4, which corresponds to option A.**Final Answer**boxed{A}

answer：Alright, so I have this problem about complex numbers. It asks for the necessary and sufficient conditions for the quotient of two complex numbers to be either a real number or a purely imaginary number. Hmm, okay, let me break this down step by step.First, let's recall what a complex number is. A complex number can be written in the form ( z = a + bi ), where ( a ) and ( b ) are real numbers, and ( i ) is the imaginary unit with the property that ( i^2 = -1 ). So, if I have two complex numbers, say ( z_1 = a + bi ) and ( z_2 = c + di ), I need to find when their quotient ( frac{z_1}{z_2} ) is real or purely imaginary.Let me start with part a) where the quotient is a real number. To find this, I think I need to express the quotient in a standard form and then see under what conditions the imaginary part becomes zero.So, let's compute ( frac{z_1}{z_2} ). To do this, I remember that I can multiply the numerator and the denominator by the conjugate of the denominator to rationalize it. The conjugate of ( z_2 = c + di ) is ( overline{z_2} = c - di ). So, multiplying both numerator and denominator by ( overline{z_2} ), we get:[frac{z_1}{z_2} = frac{a + bi}{c + di} times frac{c - di}{c - di} = frac{(a + bi)(c - di)}{(c + di)(c - di)}]Now, let's compute the denominator first:[(c + di)(c - di) = c^2 - (di)^2 = c^2 - d^2i^2 = c^2 + d^2]Since ( i^2 = -1 ), the denominator simplifies to ( c^2 + d^2 ), which is a real number.Now, let's compute the numerator:[(a + bi)(c - di) = a cdot c + a cdot (-di) + bi cdot c + bi cdot (-di)][= ac - adi + bci - bdi^2][= ac - adi + bci + bd][= (ac + bd) + (bc - ad)i]So, the numerator is ( (ac + bd) + (bc - ad)i ).Putting it all together, the quotient is:[frac{z_1}{z_2} = frac{(ac + bd) + (bc - ad)i}{c^2 + d^2}]This can be written as:[frac{ac + bd}{c^2 + d^2} + frac{bc - ad}{c^2 + d^2}i]For this quotient to be a real number, the imaginary part must be zero. That is:[frac{bc - ad}{c^2 + d^2} = 0]Since ( c^2 + d^2 ) is always positive (unless both ( c ) and ( d ) are zero, but then ( z_2 ) would be zero, and division by zero is undefined), the numerator must be zero:[bc - ad = 0]So, the necessary and sufficient condition for ( frac{z_1}{z_2} ) to be real is ( bc - ad = 0 ).Now, moving on to part b), where the quotient is a purely imaginary number. For a complex number to be purely imaginary, its real part must be zero. So, looking back at the expression for the quotient:[frac{ac + bd}{c^2 + d^2} + frac{bc - ad}{c^2 + d^2}i]The real part is ( frac{ac + bd}{c^2 + d^2} ). For this to be zero, we need:[ac + bd = 0]So, the necessary and sufficient condition for ( frac{z_1}{z_2} ) to be purely imaginary is ( ac + bd = 0 ).Wait, let me double-check that. If ( ac + bd = 0 ), then the real part is zero, and the imaginary part is ( frac{bc - ad}{c^2 + d^2}i ). So, as long as ( bc - ad ) is not zero, the quotient is purely imaginary. But if ( bc - ad = 0 ) as well, then the quotient would be zero, which is technically both real and imaginary. Hmm, but I think in the context of this problem, we're considering non-zero purely imaginary numbers, so we might need to exclude the case where both ( ac + bd = 0 ) and ( bc - ad = 0 ). But I'm not sure if that's necessary here. Maybe it's sufficient to just state ( ac + bd = 0 ) as the condition.Also, I should consider if there are any special cases or if I missed something. For example, if ( z_2 = 0 ), then the quotient is undefined, so we need to assume ( z_2 neq 0 ). Also, if ( z_1 = 0 ), then the quotient is zero, which is real, but not purely imaginary unless we consider zero as both. But I think in most contexts, zero is considered real, not purely imaginary.Another thing to consider is if ( z_1 ) and ( z_2 ) are related in some way, like being scalar multiples or something. If ( z_1 ) is a real multiple of ( z_2 ), then their quotient is real. Similarly, if ( z_1 ) is a purely imaginary multiple of ( z_2 ), then their quotient is purely imaginary. But I think the conditions I derived cover that.Let me summarize:- For ( frac{z_1}{z_2} ) to be real: ( bc - ad = 0 )- For ( frac{z_1}{z_2} ) to be purely imaginary: ( ac + bd = 0 )These conditions seem to make sense. I think I've covered all the necessary steps and checked for potential issues. I don't see any mistakes in my reasoning, so I'm confident with these results.

answer：First, I need to determine the percentage increase each time the book's price is raised successively twice, resulting in a total increase of 32.25%.I'll start by letting the original price of the book be ( P ) and the percentage increase each time be ( x % ).After the first increase, the new price becomes ( P(1 + frac{x}{100}) ).After the second increase, the price becomes ( P(1 + frac{x}{100})^2 ).According to the problem, the final price is 32.25% higher than the original price, so I can set up the equation:[ P(1 + frac{x}{100})^2 = P(1 + 0.3225) ]Simplifying this, I get:[ (1 + frac{x}{100})^2 = 1.3225 ]Taking the square root of both sides:[ 1 + frac{x}{100} = sqrt{1.3225} ]Calculating the square root of 1.3225 gives approximately 1.15, so:[ frac{x}{100} = 1.15 - 1 ][ frac{x}{100} = 0.15 ]Multiplying both sides by 100:[ x = 15 % ]Therefore, the percentage increase each time is 15%.

answer：First, I need to understand the structure of the figure ( XYZWVUT ) and identify all the sides that contribute to its perimeter. The figure is composed of three equilateral triangles: ( XYZ ), ( XWV ), and ( VUT ). Points ( W ) and ( T ) are midpoints of ( overline{XZ} ) and ( overline{XV} ), respectively.Given that ( XY = 6 ), and all triangles are equilateral, each side of these triangles is also 6 units long. 1. **Triangle ( XYZ )**: - Sides: ( XY = YZ = XZ = 6 )2. **Triangle ( XWV )**: - Since ( W ) is the midpoint of ( overline{XZ} ), ( XW = WZ = 3 ). - As ( XWV ) is equilateral, ( XV = WV = 6 ).3. **Triangle ( VUT )**: - Since ( T ) is the midpoint of ( overline{XV} ), ( VT = TX = 3 ). - As ( VUT ) is equilateral, ( UT = TV = 3 ).Next, I will list all the sides that form the perimeter of the figure ( XYZWVUT ):- ( XY = 6 )- ( YZ = 6 )- ( ZW = 3 )- ( WV = 6 )- ( VT = 3 )- ( TU = 3 )- ( UX = 3 )Adding these lengths together:[6 + 6 + 3 + 6 + 3 + 3 + 3 = 30]

answer：Okay, so I have this inequality to solve: sin ^{2018} x + cos ^{-2019} x leqslant cos ^{2018} x + sin ^{-2019} xAnd I need to find all solutions within the segment left[-frac{5 pi}{4} ; frac{3 pi}{4}right]. Hmm, that's a pretty big range, almost two full periods of the sine and cosine functions. The exponents are also quite large, 2018 and -2019, which makes me think that the behavior of sine and cosine will be significantly affected by these exponents.First, I should recall that both sine and cosine functions have outputs ranging between -1 and 1. However, when raised to even exponents like 2018, both positive and negative values will become positive. On the other hand, the negative exponents, like -2019, will invert the functions. So, cos^{-2019}x is the same as 1/cos^{2019}x, but since the exponent is odd, the sign will be preserved. Similarly, sin^{-2019}x is 1/sin^{2019}x.Wait, but hold on, if sine or cosine is zero, those terms would be undefined because we can't divide by zero. So, I need to be careful about the points where sine or cosine is zero. In the interval left[-frac{5 pi}{4} ; frac{3 pi}{4}right], sine is zero at multiples of pi, so at 0, pi, -pi, etc. Cosine is zero at odd multiples of frac{pi}{2}, so at -frac{3pi}{2}, -frac{pi}{2}, frac{pi}{2}, etc. But within our interval, cosine is zero at -frac{3pi}{2} is outside, so only at -frac{pi}{2} and frac{pi}{2}. So, I need to exclude these points from my solution set because the original inequality would be undefined there.So, my domain is left[-frac{5 pi}{4} ; frac{3 pi}{4}right] excluding -frac{pi}{2} and frac{pi}{2}. Now, looking at the inequality again:sin ^{2018} x + cos ^{-2019} x leqslant cos ^{2018} x + sin ^{-2019} xLet me rearrange the terms to bring similar terms together:sin ^{2018} x - cos ^{2018} x leqslant sin ^{-2019} x - cos ^{-2019} xHmm, that might help. Let me denote A = sin x and B = cos x to make it simpler:A^{2018} - B^{2018} leqslant A^{-2019} - B^{-2019}So, that's:A^{2018} - B^{2018} - A^{-2019} + B^{-2019} leqslant 0Hmm, not sure if that helps directly. Maybe I can factor this expression or find some common terms.Alternatively, perhaps I can consider the function f(t) = t^{2018} - t^{-2019} and analyze its behavior. Let me define:f(t) = t^{2018} - t^{-2019}So, then the inequality becomes:f(sin x) leqslant f(cos x)So, if I can understand when f(sin x) leqslant f(cos x), that would solve the inequality.Let me analyze the function f(t). Let's compute its derivative to see if it's increasing or decreasing.f'(t) = 2018 t^{2017} + 2019 t^{-2020}Since t^{2017} and t^{-2020} are both positive when t > 0, and negative when t < 0, but multiplied by positive coefficients, the derivative is positive for t > 0 and negative for t < 0. So, the function f(t) is increasing for t > 0 and decreasing for t < 0.Therefore, if both sin x and cos x are positive, then f(sin x) leqslant f(cos x) implies sin x leqslant cos x because f is increasing. Similarly, if both sin x and cos x are negative, since f is decreasing, f(sin x) leqslant f(cos x) would imply sin x geqslant cos x.But wait, sine and cosine can also be positive or negative depending on the quadrant. So, I need to consider the quadrants where x lies in the interval left[-frac{5 pi}{4} ; frac{3 pi}{4}right].Let me break down the interval into quadrants:1. -frac{5pi}{4} to -pi: This is in the third quadrant where both sine and cosine are negative.2. -pi to -frac{pi}{2}: Still in the third quadrant, both negative.3. -frac{pi}{2} to 0: Fourth quadrant, sine negative, cosine positive.4. 0 to frac{pi}{2}: First quadrant, both positive.5. frac{pi}{2} to frac{3pi}{4}: Second quadrant, sine positive, cosine negative.But we have to exclude points where sine or cosine is zero, which are at -pi, -frac{pi}{2}, 0, frac{pi}{2}, and pi, but within our interval, we only exclude -frac{pi}{2} and frac{pi}{2}.So, let's analyze each segment:1. **Third Quadrant (-frac{5pi}{4} to -pi):** Both sine and cosine are negative. Since f(t) is decreasing for t < 0, f(sin x) leqslant f(cos x) implies sin x geqslant cos x. So, we need to find where sin x geqslant cos x in this interval.In the third quadrant, both sine and cosine are negative, but their magnitudes can be compared. The angle where sin x = cos x is at x = frac{5pi}{4} (since tan x = 1), but that's at the end of our interval. Wait, actually, in the third quadrant, sin x = cos x occurs at x = frac{5pi}{4}, which is the endpoint. So, in the interval -frac{5pi}{4} to -pi, which is equivalent to frac{3pi}{4} to frac{pi}{2} in positive angles, but negative. Hmm, maybe I need to think differently.Wait, actually, -frac{5pi}{4} is equivalent to frac{3pi}{4} in the positive direction, but in the negative direction, it's in the third quadrant. So, in the third quadrant, sin x = cos x occurs at x = -frac{3pi}{4}, which is within our interval. So, from -frac{5pi}{4} to -frac{3pi}{4}, sin x leqslant cos x because as we move from -frac{5pi}{4} to -frac{3pi}{4}, sine increases from -frac{sqrt{2}}{2} to -frac{sqrt{2}}{2}? Wait, no, actually, at -frac{5pi}{4}, sine is -frac{sqrt{2}}{2}, and at -frac{3pi}{4}, sine is also -frac{sqrt{2}}{2}. Wait, that can't be right.Wait, no, let's compute specific values. At x = -frac{5pi}{4}, which is equivalent to frac{3pi}{4} in the positive direction, but in the negative direction, it's in the third quadrant. So, sine is -frac{sqrt{2}}{2} and cosine is also -frac{sqrt{2}}{2}. Wait, no, at -frac{5pi}{4}, sine is -frac{sqrt{2}}{2} and cosine is -frac{sqrt{2}}{2} as well. Wait, that's not correct because at -frac{5pi}{4}, which is 225 degrees in the negative direction, both sine and cosine are negative, but their magnitudes are equal. So, actually, at -frac{5pi}{4}, sin x = cos x = -frac{sqrt{2}}{2}. Then, as we move towards -pi, which is 180 degrees, sine becomes 0 and cosine becomes -1. So, in this interval, from -frac{5pi}{4} to -pi, sine goes from -frac{sqrt{2}}{2} to 0, and cosine goes from -frac{sqrt{2}}{2} to -1.So, in this interval, sine is increasing from -frac{sqrt{2}}{2} to 0, and cosine is decreasing from -frac{sqrt{2}}{2} to -1. Therefore, in this interval, sin x is greater than cos x because sine is increasing towards 0, while cosine is decreasing towards -1. So, sin x geqslant cos x in this interval, which, since both are negative, and f is decreasing, implies f(sin x) leqslant f(cos x). Therefore, the inequality holds for all x in -frac{5pi}{4} to -pi.2. **Third Quadrant (-pi to -frac{pi}{2}):** Both sine and cosine are negative. Again, we need to check where sin x geqslant cos x because f is decreasing.At x = -pi, sine is 0 and cosine is -1. At x = -frac{pi}{2}, sine is -1 and cosine is 0. So, in this interval, sine goes from 0 to -1, and cosine goes from -1 to 0. So, in this interval, sine is less than cosine because sine starts at 0 and goes down to -1, while cosine starts at -1 and goes up to 0. Therefore, sin x leqslant cos x in this interval, which, since both are negative, and f is decreasing, implies f(sin x) geqslant f(cos x). Therefore, the inequality does not hold in this interval.3. **Fourth Quadrant (-frac{pi}{2} to 0):** Sine is negative, cosine is positive. So, sin x is negative, and cos x is positive. Therefore, f(sin x) is (-|sin x|)^{2018} - (-|sin x|)^{-2019} = |sin x|^{2018} - (-1)^{-2019}|sin x|^{-2019} = |sin x|^{2018} + |sin x|^{-2019} because (-1)^{-2019} = -1. Similarly, f(cos x) is |cos x|^{2018} - |cos x|^{-2019}.So, the inequality becomes:|sin x|^{2018} + |sin x|^{-2019} leqslant |cos x|^{2018} - |cos x|^{-2019}Hmm, that seems complicated. Let me denote A = |sin x| and B = |cos x|, so the inequality is:A^{2018} + A^{-2019} leqslant B^{2018} - B^{-2019}But since A and B are both positive and less than or equal to 1, because they are absolute values of sine and cosine. So, A^{2018} and B^{2018} are very small because they are numbers less than 1 raised to a high power. On the other hand, A^{-2019} and B^{-2019} are very large because they are reciprocals of small numbers raised to a high power.So, let's approximate. Since A and B are between 0 and 1, A^{2018} is negligible compared to A^{-2019}, and similarly for B. So, approximately, the inequality becomes:A^{-2019} leqslant -B^{-2019}But A^{-2019} is positive, and -B^{-2019} is negative. So, a positive number is less than or equal to a negative number? That can't be true. Therefore, the inequality does not hold in this interval.Wait, but let me check that approximation. Maybe I was too hasty. Let me see:If A is very small, say approaching 0, then A^{2018} approaches 0, and A^{-2019} approaches infinity. Similarly, B is approaching 1 as x approaches 0, so B^{2018} approaches 1, and B^{-2019} approaches 1. So, the left side is approximately infinity, and the right side is approximately 1 - 1 = 0. So, infinity is not less than or equal to 0. Therefore, the inequality does not hold near x = 0.On the other hand, near x = -frac{pi}{2}, A = |sin x| approaches 1, and B = |cos x| approaches 0. So, A^{2018} approaches 1, A^{-2019} approaches 1, B^{2018} approaches 0, and B^{-2019} approaches infinity. So, the left side is approximately 1 + 1 = 2, and the right side is approximately 0 - infinity = -infinity. So, 2 is not less than or equal to -infinity. Therefore, the inequality does not hold near x = -frac{pi}{2} either.Therefore, in the entire interval -frac{pi}{2} to 0, the inequality does not hold.4. **First Quadrant (0 to frac{pi}{2}):** Both sine and cosine are positive. Since f(t) is increasing for t > 0, the inequality f(sin x) leqslant f(cos x) implies sin x leqslant cos x.So, we need to find where sin x leqslant cos x in this interval. That occurs when x leqslant frac{pi}{4} because at x = frac{pi}{4}, sin x = cos x = frac{sqrt{2}}{2}, and for x < frac{pi}{4}, sin x < cos x, and for x > frac{pi}{4}, sin x > cos x.Therefore, in the interval 0 to frac{pi}{4}, the inequality holds, and in frac{pi}{4} to frac{pi}{2}, it does not.5. **Second Quadrant (frac{pi}{2} to frac{3pi}{4}):** Sine is positive, cosine is negative. So, sin x is positive, and cos x is negative. Therefore, f(sin x) = (sin x)^{2018} - (sin x)^{-2019}, which is positive because both terms are positive. On the other hand, f(cos x) = (cos x)^{2018} - (cos x)^{-2019}. Since cos x is negative, (cos x)^{2018} is positive, and (cos x)^{-2019} is negative because it's an odd power. So, f(cos x) = (cos x)^{2018} - (cos x)^{-2019} = (cos x)^{2018} + |cos x|^{-2019} because (cos x)^{-2019} is negative.So, the inequality becomes:(sin x)^{2018} - (sin x)^{-2019} leqslant (cos x)^{2018} + |cos x|^{-2019}Again, since sin x is positive and less than or equal to 1, (sin x)^{2018} is very small, and (sin x)^{-2019} is very large. Similarly, cos x is negative, so (cos x)^{2018} is positive and small, and |cos x|^{-2019} is positive and large.So, the left side is approximately -(sin x)^{-2019}, which is a large negative number, and the right side is approximately |cos x|^{-2019}, which is a large positive number. So, a large negative number is less than a large positive number, which is always true. Therefore, the inequality holds in this interval.Wait, let me verify that. If sin x is positive, then (sin x)^{-2019} is positive, so the left side is (sin x)^{2018} - (sin x)^{-2019}, which is a small positive minus a large positive, so it's a large negative. The right side is (cos x)^{2018} + |cos x|^{-2019}, which is a small positive plus a large positive, so it's a large positive. Therefore, the inequality is true because a large negative is less than a large positive.Therefore, in the interval frac{pi}{2} to frac{3pi}{4}, the inequality holds.Putting it all together:- In -frac{5pi}{4} to -pi, the inequality holds.- In -pi to -frac{pi}{2}, it does not hold.- In -frac{pi}{2} to 0, it does not hold.- In 0 to frac{pi}{4}, it holds.- In frac{pi}{4} to frac{pi}{2}, it does not hold.- In frac{pi}{2} to frac{3pi}{4}, it holds.Therefore, the solution set is:left[-frac{5pi}{4}, -piright) cup left(0, frac{pi}{4}right] cup left(frac{pi}{2}, frac{3pi}{4}right]Wait, but in the interval -frac{5pi}{4} to -pi, we have to check if the endpoints are included. At x = -frac{5pi}{4}, both sine and cosine are -frac{sqrt{2}}{2}, so plugging into the original inequality:left(-frac{sqrt{2}}{2}right)^{2018} + left(-frac{sqrt{2}}{2}right)^{-2019} leqslant left(-frac{sqrt{2}}{2}right)^{2018} + left(-frac{sqrt{2}}{2}right)^{-2019}Which is equality, so x = -frac{5pi}{4} is included. At x = -pi, sine is 0 and cosine is -1. Plugging in:0^{2018} + (-1)^{-2019} leqslant (-1)^{2018} + 0^{-2019}But 0^{-2019} is undefined, so x = -pi is excluded. Similarly, at x = 0, sine is 0, so 0^{-2019} is undefined, so x = 0 is excluded. At x = frac{pi}{4}, sine and cosine are equal, so plugging in:left(frac{sqrt{2}}{2}right)^{2018} + left(frac{sqrt{2}}{2}right)^{-2019} leqslant left(frac{sqrt{2}}{2}right)^{2018} + left(frac{sqrt{2}}{2}right)^{-2019}Which is equality, so x = frac{pi}{4} is included. At x = frac{pi}{2}, cosine is 0, so 0^{-2019} is undefined, so x = frac{pi}{2} is excluded. At x = frac{3pi}{4}, sine is frac{sqrt{2}}{2} and cosine is -frac{sqrt{2}}{2}. Plugging in:left(frac{sqrt{2}}{2}right)^{2018} + left(-frac{sqrt{2}}{2}right)^{-2019} leqslant left(-frac{sqrt{2}}{2}right)^{2018} + left(frac{sqrt{2}}{2}right)^{-2019}Simplifying:left(frac{sqrt{2}}{2}right)^{2018} + left(-frac{sqrt{2}}{2}right)^{-2019} = left(frac{sqrt{2}}{2}right)^{2018} - left(frac{sqrt{2}}{2}right)^{-2019}And the right side:left(frac{sqrt{2}}{2}right)^{2018} + left(frac{sqrt{2}}{2}right)^{-2019}So, the left side is less than the right side because we have a negative term on the left. Therefore, the inequality holds at x = frac{3pi}{4}.Wait, but earlier I concluded that in the interval frac{pi}{2} to frac{3pi}{4}, the inequality holds, and since x = frac{3pi}{4} is included, that's correct.So, to summarize, the solution set is:left[-frac{5pi}{4}, -piright) cup left(0, frac{pi}{4}right] cup left(frac{pi}{2}, frac{3pi}{4}right]But I need to check if x = frac{3pi}{4} is included. As I saw earlier, at x = frac{3pi}{4}, the inequality holds, so it should be included. Similarly, x = -frac{5pi}{4} is included because it's a point of equality.Therefore, the final solution is:left[-frac{5pi}{4}, -piright) cup left(0, frac{pi}{4}right] cup left(frac{pi}{2}, frac{3pi}{4}right]