answer:: 1. **Determine the Domain of Validity (ODZ):** We must ensure that the expressions in the equation are defined: [ left{ begin{array}{l} sin 2t neq 0, sin t neq pm frac{1}{2} end{array} right. ] 2. **Simplify the left-hand side using a trigonometric identity:** We know the identity for the difference of sines: [ sin alpha - sin beta = 2 cos left( frac{alpha + beta}{2} right) sin left( frac{alpha - beta}{2} right) ] Applying it to (sin 3t - sin t): [ sin 3t - sin t = 2 sin t cos 2t ] So, the equation becomes: [ 2 sin t cos 2t = frac{8 cos t cot 2t}{4 - sin^{-2} t} ] 3. **Simplify the right-hand side:** Express (cot 2t) and compute: [ cot 2t = frac{cos 2t}{sin 2t} ] Substituting (cot 2t): [ frac{8 cos t cos 2t}{sin 2t (4 - sin^{-2} t)} ] 4. **Combine and rearrange the terms:** Equate the two sides: [ 2 sin t cos 2t = frac{8 cos t cos 2t}{sin 2t (4 - sin^{-2} t)} ] Simplify the denominator: [ 4 - sin^{-2} t = frac{4 sin^2 t - 1}{sin^2 t} ] Substitute back: [ 2 sin t cos 2t = frac{8 cos t cos 2t sin^2 t}{2 sin t cos t (4 sin^2 t - 1)} ] 5. **Further simplification:** Cancel out common terms: [ 2 sin t cos 2t = 2 sin t cos 2t cdot frac{2}{4 sin^2 t - 1} ] Combine and solve: [ 2 sin t cos 2t left(1 - frac{2}{4 sin^2 t - 1}right) = 0 ] 6. **Solve the resulting equation:** The product of factors equals zero, so either factor can be zero: - If (cos 2t = 0): [ 2t = frac{pi}{2} + pi k quad implies quad t_1 = frac{pi}{4} + frac{pi k}{2} = frac{pi}{4}(2k+1), quad k in mathbb{Z} ] - If (1 - frac{2}{4 sin^2 t - 1} = 0): [ frac{2}{4 sin^2 t - 1} = 1 quad implies quad 4 sin^2 t - 1 = 2 quad implies quad 4 sin^2 t = 3 quad implies quad sin^2 t = frac{3}{4} ] Thus: [ sin t = pm frac{sqrt{3}}{2} quad implies quad t_2 = pm frac{pi}{3} + pi l, quad l in mathbb{Z} ] **Conclusion:** The solutions are: [ t_1 = frac{pi}{4}(2k+1), quad t_2 = pm frac{pi}{3} + pi l quad text{where} quad k, l in mathbb{Z} ] [ boxed{t_1 = frac{pi}{4}(2k+1), t_2 = pm frac{pi}{3} + pi l text{ where } k, l in mathbb{Z}} ]

answer:Oliver rode the ferris wheel 5 times and the bumper cars 4 times, so he went on a total of 5 + 4 = 9 rides. Each ride cost 7 tickets, so for 9 rides, he would have used 9 * 7 = boxed{63} tickets.

answer:Given the conditions, when x in (-infty, 0), we have the inequality xf'(x) - 2f(x) > 0. To analyze this, let's define a new function g(x) = frac{f(x)}{x^{2}}. The derivative of g(x) with respect to x can be calculated as follows: [ g'(x) = left(frac{f(x)}{x^{2}}right)' = frac{x^{2}f'(x) - 2xf(x)}{x^{4}} = frac{x f'(x) - 2f(x)}{x^{3}} ] Given the original inequality xf'(x) - 2f(x) > 0, we substitute into the derivative of g(x) to get: [ g'(x) = frac{x f'(x) - 2f(x)}{x^{3}} < 0 ] This indicates that g(x) is monotonically decreasing on its domain (-infty, 0). Now, consider the inequality f(x+2023) - (x+2023)^{2}f(-1) < 0. We can rewrite this inequality in terms of g(x): [ frac{f(x+2023)}{(x+2023)^{2}} < frac{f(-1)}{(-1)^{2}} ] This simplifies to: [ g(x+2023) < g(-1) ] Given that g(x) is monotonically decreasing, for g(x+2023) < g(-1) to hold, we must have: [ left{begin{array}{l} x+2023 > -1 x+2023 < 0 end{array}right. ] Solving these inequalities gives us the solution set for x: [ -2024 < x < -2023 ] Therefore, the correct answer is boxed{A}.

answer:First, we convert the numbers from base 6 to base 10: [ 5_6 = 5_{10} ] [ 21_6 = 2 cdot 6^1 + 1 cdot 6^0 = 12 + 1 = 13_{10} ] Now, add these numbers in base 10: [ 5_{10} + 13_{10} = 18_{10} ] Next, we convert the sum back to base 6. To do this, we find the largest power of 6 less than 18, which is 6^1 = 6, and determine how many times 6 fits into 18, which is 3 times. The remainder is 0. Therefore, [ 18_{10} = 30_6 ] Thus, the sum of 5_6 + 21_6 in base 6 is boxed{30_6}.