The number of solutions of the equation sin^{ | vedclass

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(C) Given equation: $\sin^{65}x - \cos^{65}x = -1$.Since $-1 \le \sin x \le 1$ and $-1 \le \cos x \le 1$,we have $-1 \le \sin^{65}x \le 1$ and $-1 \le

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$2$

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(C) Given equation: $\sin^{65}x - \cos^{65}x = -1$.Since $-1 \le \sin x \le 1$ and $-1 \le \cos x \le 1$,we have $-1 \le \sin^{65}x \le 1$ and $-1 \le \cos^{65}x \le 1$.Let $u = \sin^{65}x$ and $v = \cos^{65}x$. Then $u - v = -1$,which implies $v = u + 1$.Since $v \in [-1, 1]$,we have $-1 \le u + 1 \le 1$,so $-2 \le u \le 0$.Also,$u = \sin^{65}x \in [-1, 1]$,so the possible values for $u$ are in $[-1, 0]$.Case $1$: If $u = -1$,then $\sin^{65}x = -1 \implies \sin x = -1$. Then $v = -1 + 1 = 0$,so $\cos^{65}x = 0 \implies \cos x = 0$.For $x \in (-\pi, \pi)$,$\sin x = -1$ gives $x = -\pi/2$. At $x = -\pi/2$,$\cos x = 0$,which satisfies $\cos^{65}x = 0$. Thus,$x = -\pi/2$ is a solution.Case $2$: If $u = 0$,then $\sin^{65}x = 0 \implies \sin x = 0$. Then $v = 0 + 1 = 1$,so $\cos^{65}x = 1 \implies \cos x = 1$.For $x \in (-\pi, \pi)$,$\sin x = 0$ gives $x = 0$. At $x = 0$,$\cos x = 1$,which satisfies $\cos^{65}x = 1$. Thus,$x = 0$ is a solution.Therefore,the solutions are $x = -\pi/2$ and $x = 0$. The total number of solutions is $2$.

$List-I$	$List-II$
$(I)$ $\{x \in[-\frac{2 \pi}{3}, \frac{2 \pi}{3}]: \cos x+\sin x=1\}$	$(P)$ has two elements
$(II)$ $\{x \in[-\frac{5 \pi}{18}, \frac{5 \pi}{18}]: \sqrt{3} \tan 3 x=1\}$	$(Q)$ has three elements
$(III)$ $\{x \in[-\frac{6 \pi}{5}, \frac{6 \pi}{5}]: 2 \cos (2 x)=\sqrt{3}\}$	$(R)$ has four elements
$(IV)$ $\{x \in[-\frac{7 \pi}{4}, \frac{7 \pi}{4}]: \sin x-\cos x=1\}$	$(S)$ has five elements
	$(T)$ has six elements

The number of solutions of the equation $\sin^{65}x - \cos^{65}x = -1$ for $x \in (-\pi, \pi)$ is:

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