Question

The number of solutions of the equation $\left|cotx\right|=cotx+\frac{1}{sinx},\left(0\le x\le 2\pi \right)$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (C)

(i) When $x\in \left[0,\frac{\pi }{2}\right]\cup \left(\pi ,\frac{3\pi }{2}\right)$ then $\cot x\ge 0$
$\Rightarrow \cot x=\cot x+\frac{1}{\sin x}\Rightarrow \frac{1}{\sin x}=0$
$\Rightarrow$ No solution exist
(ii) When $x\in \left(\frac{\pi }{2},\pi \right)\cup \left(\frac{3\pi }{2},2\pi \right)$ then $\cot x<0$
$\therefore -\cot x=\cot x+\frac{1}{\sin x}$
$\Rightarrow -2\cot x=\frac{1}{\sin x}$
$\Rightarrow \cos x=\frac{-1}{2}\Rightarrow x=\frac{2\pi }{3}$