Giải phương trình tanx = cotx.

ĐKXĐ: $\left\{ {\begin{array}{*{20}{c}}{\sin x \ne 0}\\{\cos x \ne 0}\end{array}} \right. \Rightarrow \sin 2x \ne 0 \Leftrightarrow 2x \ne k\pi \Leftrightarrow x \ne \frac{{k\pi }}{2}\left( {k \in \mathbb{Z}} \right)$

tanx = cotx $\Leftrightarrow \tan x = \frac{1}{{\tan x}} \Leftrightarrow {\tan ^2}x = 1$

\( \Leftrightarrow \left[ { $$\begin{array}{*{20}{c}}{\tan x = 1}\\{\tan x = - 1}\end{array}$$ } \right. \Leftrightarrow \left[ { $$\begin{array}{*{20}{c}}{x = \frac{\pi }{4} + k\pi }\\{x = - \frac{\pi }{4} + k\pi }\end{array}$$ } \right. \Rightarrow x = \frac{\pi }{4} + k\frac{\pi }{2}\left( {k \in \mathbb{Z}} \right)\).