Official solution
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