Giải PTLG cot2x = $\frac{1 – tanx}{1 + tanx}$

$\begin{array}{l} \dfrac{{1 – \tan x}}{{1 + \tan x}} = \cot 2x\\  \Leftrightarrow \dfrac{{\tan \dfrac{\pi }{4} – \tan x}}{{1 + \tan \dfrac{\pi }{4}\tan x}} = \cot 2x\\  \Leftrightarrow \tan \left( {\dfrac{\pi }{4} – x} \right) = \cot 2x\\  \Leftrightarrow \tan \left( {\dfrac{\pi }{2} – \left( {\dfrac{\pi }{4} + x} \right)} \right) = \cot 2x\\  \Leftrightarrow \cot \left( {\dfrac{\pi }{4} + x} \right) = \cot 2x\\  \Rightarrow 2x = x + \dfrac{\pi }{4} + k\pi \\  \Leftrightarrow x = \dfrac{\pi }{4} + k\pi \left( {k \in \mathbb{Z}} \right) \end{array}$