The number of solutions of the equation | x| = cot x + cosecx in...

If $x$ lies in ${1}^{st} or {3}^{rd}$ quadrant then \left|\cot x\right| = cot x,$thus equation becomes<br />$cot x = cot x + \dfrac{1}{sin x} ⇒ \dfrac{1}{sin x} = 0,$not possible<br />If$x$lies in${2}^{nd} or {4}^{th}$quadrant, then \left|\cot x\right| = - cot x,$ thus equation becomes
$- cot x = cot x + \dfrac{1}{sin x}$
$⇒ - 2cot x = \dfrac{1}{sin x} ⇒ - 2cos xsin x = sin x$
$⇒ sin x\left(1 + 2cos x\right) = 0 ⇒ cos x = - \dfrac{1}{2}\left(∵ sin x \ne 0\right)$
$⇒ x = \dfrac{2\pi}{3}$ is the only solution
(In ${2}^{nd}$ quad.)
In \left[0,2\pi\right]$there is only one solution<br />In \left[0,10\pi\right]$, we get, five solutions