Question

The numbers of solutions of the equation $|\cot x|=\cot x+\frac{1}{\sin x}$ in the interval $[0,2\pi ]$ is

Answer 1

Case $1:x\in [0,\frac{\pi }{2}]\cup [0,\frac{3\pi }{2}]$
$\cot x=\cot x+\frac{1}{\sin x}\Rightarrow$ not possible
Case $2:x\in [\frac{\pi }{2},\pi ]\cup [\frac{3\pi }{2},2\pi ]$
$-\cot x=\cot x+\frac{1}{\sin x}$
$\Rightarrow \frac{-2\cos x}{\sin x}=\frac{1}{\sin x}$
$\Rightarrow \cos x=\frac{-1}{2}$
$\Rightarrow x=\frac{2\pi }{3},\frac{4\pi }{3}$
$=1$