$\cot x -\tan x -\cot 2x =1$
$\Leftrightarrow \dfrac{\cos x}{\sin x}-\dfrac{\sin x}{\cos x}-\cot 2x =1$
$\Leftrightarrow \dfrac{\cos^2 x-\sin^2 x}{\sin x \cos x}-\cot 2x =1$
$\Leftrightarrow \dfrac{\cos 2x}{\sin x \cos x}-\cot 2x =1$
$\Leftrightarrow \dfrac{2\cos 2x}{2\sin x \cos x}-\cot 2x =1$
$\Leftrightarrow \dfrac{2\cos 2x}{\sin 2x}-\cot 2x =1$
$\Leftrightarrow 2\cot 2x -\cot 2x =1$ $\Leftrightarrow \cot 2x =1 \Leftrightarrow 2x =\dfrac{\pi}{4}+k\pi \Leftrightarrow x= \dfrac{\pi}{8}+\dfrac{k\pi}{2};\ k\in Z$