Question:
If $0 \leq x<\frac{\pi}{2}$, then the number of values of $x$ for which $\sin x-\sin 2 x+\sin 3 x=0$, is
Correct Option: 1
Solution:
$\sin x-\sin 2 x+\sin 3 x=0$
$\Rightarrow(\sin x+\sin 3 x)-\sin 2 x=0$
$\Rightarrow 2 \sin x \cdot \cos x-\sin 2 x=0$
$\Rightarrow \sin 2 x(2 \cos x-1)=0$
$\Rightarrow \sin 2 x=0$ or $\cos x=\frac{1}{2}$
$\Rightarrow x=0, \frac{\pi}{3}$