Explanation

(c)
Given equation is
$\sin{x\cos{3x = \sin{3x\cos{5x}}}}$
$\Rightarrow \ \ \ 2\sin{x\cos{3x - 2\sin{3x\cos{5x = 0}}}}$
$\Rightarrow \ \sin{(3x + x) - \sin{(3x - x) - \sin{(3x + 5x)}}} + \sin{(5x - 3x) = 0}$
$\Rightarrow \ \sin{4x - \sin{2x - \sin{8x + \sin{2x = 0}}}}$
$\sin{4x - \sin{8x} = 0}$
$\Rightarrow \ \ 2\cos{\left( \frac{4x + 8x}{2} \right)\sin{\left( \frac{8x - 4x}{2} \right) = 0}}$
$\Rightarrow \ 2\cos{6x\sin{2x = 0}}$
$\Rightarrow \ \cos{6x = 0}\ \ \$or$\ \ \ \ \sin\ 2x = 0$
$\Rightarrow \ \ \ 6x = (2n + 1)\frac{\pi}{2}\$ or $x = \frac{n\pi}{2}$
$\Rightarrow \ \ x = (2n + 1)\frac{\pi}{12}\$ or $x = \frac{n\pi}{2}$
$\Rightarrow \ \ \ x = 0,\frac{\pi}{2},\ \frac{\pi}{12},\ \frac{3\pi}{12},\ \frac{5\pi}{12} \in \left\lbrack 0,\ \frac{\pi}{2} \right\rbrack$
$\therefore$ Number of solutions is 5