Solve: $\;$ $2 \tan^{-1} \left(\cos x\right) = \tan^{-1} \left(2 \text{ cosec }x\right)$
$2 \tan^{-1} \left(\cos x\right) = \tan^{-1} \left(2 \text{ cosec }x\right)$
i.e. $\;$ $\tan^{-1} \left(\dfrac{2 \cos x}{1 - \cos^2 x}\right) = \tan^{-1} \left(2 \text{ cosec } x\right)$ $\;\;\;$ $\left[\text{Note: } 2 \tan^{-1} x = \tan^{-1} \left(\dfrac{2 x}{1 - x^2}\right)\right]$
i.e. $\;$ $\dfrac{2 \cos x}{1 - \cos^2 x} = 2 \text{ cosec }x$ $\;\;\;$ $\left[\text{Note: } \tan^{-1} \text{ is one - one}\right]$
i.e. $\;$ $\dfrac{\cos x}{\sin^2 x} = \dfrac{1}{\sin x}$
i.e. $\;$ $\dfrac{\cos x}{\sin x} = 1$ $\;\;\;$ $\left[\text{provided } \sin x \neq 0\right]$
i.e. $\;$ $\tan x = 1$
i.e. $\;$ $x = \tan^{-1} \left(1\right) = \dfrac{\pi}{4}$
Verification:
Putting $x = \dfrac{\pi}{4}$ in the given equation,
$LHS = 2 \tan^{-1} \left[\cos \left(\dfrac{\pi}{4}\right)\right]$
$ = 2 \tan^{-1} \left(\dfrac{1}{\sqrt{2}}\right) = \tan^{-1} \left[\dfrac{2 \times \dfrac{1}{\sqrt{2}}}{1 - \dfrac{1}{2}}\right] = \tan^{-1} \left(2 \sqrt{2}\right)$
$RHS = \tan^{-1} \left[2 \text{ cosec } \left(\dfrac{\pi}{4}\right)\right] = \tan^{-1} \left(2 \sqrt{2}\right) = LHS$
$\therefore \;$ The solution set is $\left\{\dfrac{\pi}{4} \right\}$