Question

Number of solutions of the equation
$|\sin x|=\sin x-2\cos x$ in $[0,4\pi ]$ is

Answer 1

Case 1: $\sin x\ge 0$
$\sin x=\sin x-2\cos x\Rightarrow \cos x=0$
$\Rightarrow x=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\frac{7\pi }{2}$
Since, $\sin x\ge 0\Rightarrow x=\frac{\pi }{2},\frac{5\pi }{2}$

Case 2: $\sin x<0$
$-\sin x=\sin x-2\cos x$
$\Rightarrow -2\sin x=-2\cos x\Rightarrow \tan x=1$
$\Rightarrow x=\frac{\pi }{4},\frac{5\pi }{4},\frac{9\pi }{4},\frac{13\pi }{4}$
Since, $\sin x<0\Rightarrow x=\frac{5\pi }{4},\frac{13\pi }{4}$
$\therefore$ Number of solutions is $4$.