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

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

$$\begin{gathered} Case\left(i\right): Let cotx\ge 0. Then \\ cotx=cotx+\frac{1}{\sin x} \\ \frac{1}{\sin x}=0 \\ \sin x=\infty which is impossible \end{gathered}$$

$$\begin{gathered} Case\left(ii\right): Let cotx<0. Then \\ -cotx=cotx+\frac{1}{\sin x} \end{gathered}$$

$$\begin{gathered} \Rightarrow \cos x=-\frac{1}{2} \\ \Rightarrow x=\frac{2\mathrm{\pi }}{3},\frac{4\mathrm{\pi }}{3} \left(\because 0<x<2\mathrm{\pi }\right) \\ \Rightarrow x=\frac{2\mathrm{\pi }}{3} \left(\because only cotx<0 has solution\right) \end{gathered}$$