Solve the following equation: $2{\tan }^{-1}\left(\cos x\right)={\tan }^{-1}\left(2\text{cosec}x\right)$

Given:
$2{\tan }^{-1}\left(\cos x\right)={\tan }^{-1}\left(2\text{cosec}x\right)$
$\{2{\tan }^{-1}x={\tan }^{-1}\frac{2x}{1-x^2}\}$
$\Rightarrow {\tan }^{-1}\frac{2\cos x}{1-{\cos }^{2}x}={\tan }^{-1}\left(2\text{cosec}x\right)$
$\Rightarrow {\tan }^{-1}\left(\frac{2\cos x}{{\sin }^{2}x}\right)={\tan }^{-1}\left(2\text{cosec}x\right)$
$\Rightarrow \frac{2\cos x}{{\sin }^{2}x}=2\text{cosec}x$
$\Rightarrow \frac{\cos x}{{\sin }^{2}x}=\text{cosec}x$
$\Rightarrow \frac{\cos x}{{\sin }^{2}x}=\frac{1}{\sin x}$
$\Rightarrow \cos x=\frac{{\sin }^{2}x}{\sin x}$
$\Rightarrow \cos x=\sin x$
$\Rightarrow 1=\frac{\sin x}{\cos x}$
$\Rightarrow \tan x=1$
$\Rightarrow \tan x=\tan \frac{\pi }{4}$
$\Rightarrow x=n\pi +\frac{\pi }{4},n\in Z$
Hence, the value of $x$ is $n\pi +\frac{\pi }{4},n\in Z$.