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

$\mid \cot x \mid=\cot x+\cfrac{1}{\sin x} \,\,\,\,\,\,\, 0<x<2\pi$

$\mid \cot x \mid$ is positive if $\cot x>0$

Then,

$\cot x=\cot x+\cfrac{1}{\sin x}$

or, $\cfrac{1}{\sin x}=0$

or, $\sin x=\infty$ which is impossible.

$\mid \cot x \mid$ is -ve if $\cot x<0$

$-\cot x=\cot x+\cfrac{1}{\sin x}$

or,$-2\cot x=\cfrac{1}{\sin x}$

or, $\cfrac{-2\cos x}{\sin x}=\cfrac{1}{\sin x}$

$\sin x \ne 0$ because $x \ne 0$

or, $-2\cos x=1$

or, $\cos x=\cfrac{-1}{2}=\cos \cfrac{2\pi}{3}$

or, $x=2n\pi \pm \cfrac{2\pi}{3}$

If $n=1$

$x=2\pi-\cfrac{2\pi}{3}=\cfrac{4\pi}{3}$

which is in the range $(0,2\pi)$

$\therefore x=\cfrac{2\pi}{3},\cfrac{4\pi}{3}$

$\therefore n=2$