If $0\le x<\frac{\pi }{2}$ ,then the number of values of x for which  $\sin x-\sin 2x+\sin 3x=0$ , is 

Solution:

we have,

$\begin{array}{l}\sin x-\sin 2x+\sin 3x=0\\ \Rightarrow (\sin 3x+\sin x)-\sin 2x=0\\ \Rightarrow 2\sin 2x\cos x-\sin 2x=0\\ \Rightarrow \sin 2x(2\cos x-1)=0\\ \Rightarrow \sin 2x=0,\cos x=\frac{1}{2}\Rightarrow x=0,\frac{\pi }{3}\left[\because 0\le x<\frac{\pi }{2}\right]\end{array}$