Solve

$2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x)$

$\Rightarrow \tan ^{-1}\left(\frac{2 \cos x}{1-\cos ^{2} x}\right)=\tan ^{-1}(2 \operatorname{cosec} x)$      $\left[2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}\right]$

$\Rightarrow \frac{2 \cos x}{1-\cos ^{2} x}=2 \operatorname{cosec} x$

$\Rightarrow \frac{2 \cos x}{\sin ^{2} x}=\frac{2}{\sin x}$

$\Rightarrow \cos x=\sin x$

$\Rightarrow \tan x=1$

$\therefore x=\frac{\pi}{4}$