Number of real values of x in (0, pi) for whi | vedclass

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(B) Given the inequality: $\frac{8}{3\sin x - \sin 3x} + 3\sin^2 x \le 5$.Using the identity $\sin 3x = 3\sin x - 4\sin^3 x$,the denominator becomes $

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$1$

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(B) Given the inequality: $\frac{8}{3\sin x - \sin 3x} + 3\sin^2 x \le 5$.Using the identity $\sin 3x = 3\sin x - 4\sin^3 x$,the denominator becomes $3\sin x - (3\sin x - 4\sin^3 x) = 4\sin^3 x$.Substituting this into the inequality: $\frac{8}{4\sin^3 x} + 3\sin^2 x \le 5 \Rightarrow \frac{2}{\sin^3 x} + 3\sin^2 x \le 5$.Let $f(x) = \frac{2}{\sin^3 x} + 3\sin^2 x - 5$. We want to find $x \in (0, \pi)$ such that $f(x) \le 0$.Let $t = \sin x$. Since $x \in (0, \pi)$,$t \in (0, 1]$.The expression becomes $g(t) = \frac{2}{t^3} + 3t^2 - 5$.By the $AM$-$GM$ inequality for three terms $\frac{1}{t^3}, \frac{1}{t^3}, t^2, t^2, t^2$:$\frac{\frac{1}{t^3} + \frac{1}{t^3} + t^2 + t^2 + t^2}{5} \ge \sqrt[5]{\frac{1}{t^3} \cdot \frac{1}{t^3} \cdot t^2 \cdot t^2 \cdot t^2} = \sqrt[5]{1} = 1$.So,$\frac{2}{t^3} + 3t^2 \ge 5$.The equality holds if and only if $\frac{1}{t^3} = t^2$,which implies $t^5 = 1$,so $t = 1$.Since $\sin x = 1$ for $x \in (0, \pi)$ only at $x = \frac{\pi}{2}$,the inequality $f(x) \le 0$ holds only for $x = \frac{\pi}{2}$.Thus,there is only $1$ real value.

Number of real values of $x \in (0, \pi)$ for which $\frac{8}{3\sin x - \sin 3x} + 3\sin^2 x \le 5$ is:

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