Question

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

• A. $0$
• B. $2$
• C. $1$
• D. $3$

Answer 1

The correct option is C $1$

$|\cot x|=\{\begin{array}{ccc}\cot x& \text{if}& x\in (0,\frac{\pi }{2}]\cup (\pi ,\frac{3\pi }{2}]\\ -\cot x& \text{if}& x\in (\frac{\pi }{2},\pi )\cup (\frac{3\pi }{2},2\pi )\end{array}$

Case $1:\text{if}x\in (0,\frac{\pi }{2}]\cup (\pi ,\frac{3\pi }{2}]$
$\Rightarrow \cot x=\cot x+\text{cosec}x\Rightarrow \text{cosec}x=0$
no solution possible

Case $2:\text{if}x\in (\frac{\pi }{2},\pi )\cup (\frac{3\pi }{2},2\pi )$
$\Rightarrow -\cot x=\cot x+\frac{1}{\sin x}\Rightarrow \cos x=-\frac{1}{2}\Rightarrow x=\frac{2\pi }{3}$