If ‘n’ be the number of solutions of the equation $\left|\cot x\right|=\cot x+\frac{1}{\sin x}\left(0<x<2\pi \right)$, then $n=$

$\left|\cot x\right|=\left\{\begin{array}{l}-\cot x,\cot x<0\\ \cot x,\cot x\ge 0\end{array}\right.$

$\text{ Case I: }\cot x\ge 0$

$\cot x=\cot x+\frac{1}{\sin x}\Rightarrow \frac{1}{\sin x}=0$

$\Rightarrow \sin x=\mathrm{\infty }\text{, this is not possible }$

$$\begin{gathered} \text{ Case II: }\cot x<0\Rightarrow -\cot x=\cot x+\frac{1}{\sin x} \\ \Rightarrow \cos x=-\frac{1}{2} \end{gathered}$$

$\Rightarrow \cos x=\cos \left(\pi +\frac{\pi }{3}\right)\text{ or }\cos \left(\pi -\frac{\pi }{3}\right)$

$\therefore x=\frac{2\pi }{3},\frac{4\pi }{3}\text{ but }x=\frac{4\pi }{3}\Rightarrow \cot x>0$

$\therefore n=1$