Question

The number of solution(s) of $\sin 3x-\sin x=4{\cos }^{2}x-2$ for $x\in [0,2\pi ]$ is

• A. $1$
• B. $2$
• C. More than $2$ but finite
• D. infinitely many

Answer 1

The correct option is C More than $2$ but finite

Given, $\sin 3x-\sin x=4{\cos }^{2}x-2$
$\Rightarrow 2\cos 2x\sin x=2(2{\cos }^{2}x-1)\Rightarrow 2\cos 2x\sin x=2\cos 2x\Rightarrow 2\cos 2x(\sin x-1)=0\Rightarrow \cos 2x=0,\sin x=1$

As $x\in [0,2\pi ]\Rightarrow 2x\in [0,4\pi ]$
So, for $\cos 2x=0$
$\Rightarrow 2x=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\frac{7\pi }{2}\Rightarrow x=\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}\sin x=1\Rightarrow x=\frac{\pi }{2}$
Therefore, the number of solution of the given equation is $5$.