If the general solution set of sin x + 3 sin | vedclass

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(C) Given equation: $\sin x + \sin 5x + 3 \sin 3x = 0$ Using the formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$,we get: $2 \sin

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$\{\frac{\sqrt{3}}{2}, 0, -\frac{\sqrt{3}}{2}\}$

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(C) Given equation: $\sin x + \sin 5x + 3 \sin 3x = 0$ Using the formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$,we get: $2 \sin 3x \cos 2x + 3 \sin 3x = 0$ $\sin 3x (2 \cos 2x + 3) = 0$ Since $2 \cos 2x + 3 = 0$ implies $\cos 2x = -\frac{3}{2}$,which is impossible as $-1 \le \cos 2x \le 1$,we must have $\sin 3x = 0$. Thus,$3x = n\pi$,or $x = \frac{n\pi}{3}$ for $n \in \mathbb{Z}$. The set of values for $\sin x$ is $\{\sin(0), \sin(\frac{\pi}{3}), \sin(\frac{2\pi}{3}), \sin(\pi), \sin(\frac{4\pi}{3}), \sin(\frac{5\pi}{3})\}$. Evaluating these: $\{0, \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}, 0, -\frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2}\}$. The set of distinct values is $\{0, \frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2}\}$.

$List-I$	$List-II$
$(I)$ $\{x \in[-\frac{2 \pi}{3}, \frac{2 \pi}{3}]: \cos x+\sin x=1\}$	$(P)$ 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

If the general solution set of $\sin x + 3 \sin 3x + \sin 5x = 0$ is $S$,then $\{\sin \alpha \mid \alpha \in S\} = $

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