sin x $-$ sin 2x + sin 3x = 0 $x \in \left[ {0,{\pi \over 2}} \right)$ $\Rightarrow$ (sin3x + sinx) $-$ sin2x = 0 $\Rightarrow$ 2sin2x.cos2x $-$ sin2x = 0 $\Rightarrow$ sin2x (2cosx $-$ 1) = 0 sin 2x = 0 x = 0 and cos x = ${1 \over 2}$ and x = ${\pi \over 3}$ two solutions