Explanation

(b)
Given, $\sin{x + \sin{5x = \sin{3x}}}$
$\Rightarrow \ \ 2\sin{3x + \cos{2x = \sin{3x}}}$
$\Rightarrow \ \sin{3x\left( 2\cos{2x - 1} \right) = 0}$
$\Rightarrow \ \sin{3x = 0}$
Or $2\cos{2x - 1 = 0}$
$\Rightarrow \ \ 3x = 0,\ \pi$ or $2x = \frac{\pi}{3}$
$\Rightarrow \ \ x = 0,\ \ x = \frac{\pi}{3}$ or $x = \frac{\pi}{6}$
$\therefore$ Solutions in $\left( 0,\frac{\pi}{2} \right)$ are $\frac{\pi}{3},\frac{\pi}{6}$