Answer:
Option B
Explanation:
We have,
$\sin x$ +sin 3x+sin 5x=0
$\Rightarrow$ $\sin x+\sin 5x+\sin x$=0
$\Rightarrow$ 2$\sin 3x \cos 2x+\sin 3x$=0
$\Rightarrow$ $\sin 3x (2\cos 2x +1)$=0
$\Rightarrow$ $\sin 3x=0$ or $\cos 2x=-\frac{1}{2}=\cos \frac{2 \pi}{3}$
$\Rightarrow$ $3x=n\pi $ or $2n=2n\pi\pm\frac{2\pi}{3}$
$\Rightarrow$ $x=\frac{n\pi}{3}$ or $x=n\pi\pm\frac{\pi}{3}$
But, it is given that
$x \epsilon \left[\frac{\pi}{2},\frac{3\pi}{2}\right]$
$\therefore$ $x=\frac{2\pi}{3},\pi,\frac{4\pi}{3}$
$\therefore$ Number of solutions is 3.
