Question

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

• A. $3$
• B. $4$
• C. $2$
• D. $1$

Answer 1

The correct option is C $2$

$\sin x-\sin 2x+\sin 3x=0$
$\Rightarrow \sin x+\sin 3x-\sin 2x=0$
$\Rightarrow 2\sin 2x\cos x-\sin 2x=0$
$\Rightarrow \sin 2x(2\cos x-1)=0$

Case-$1$ :
$\sin 2x=0$
$\Rightarrow x=0$

Case-$2$ :
$\Rightarrow 2\cos x-1=0$
$\Rightarrow \cos x=\frac{1}{2}$
$\Rightarrow x=\frac{\pi }{3}$
Hence, $x=0,\frac{\pi }{3}$
$\therefore$ Number of values of $x=2$