2 câu trả lời
\[\begin{array}{l} \tan x - 2\cot x + 1 = 0\\ DK:\,\,\,\left\{ \begin{array}{l} \cos x \ne 0\\ \sin x \ne 0 \end{array} \right. \Leftrightarrow \sin 2x \ne 0.\\ \Leftrightarrow \tan x - 2.\frac{1}{{\tan x}} + 1 = 0\\ \Leftrightarrow {\tan ^2}x + \tan x - 2 = 0\\ \Leftrightarrow \left( {\tan x - 1} \right)\left( {\tan x + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \tan x - 1 = 0\\ \tan x + 2 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \tan x = 1\\ \tan x = - 2 \end{array} \right.. \end{array}\]
$$\eqalign{ & \tan x - \cot x + 1 = 0\, \cr & DKXD:\,\left\{ \matrix{ \sin x \ne 0 \hfill \cr \cos x \ne 0 \hfill \cr} \right. \Leftrightarrow \sin 2x \ne 0 \cr & \Leftrightarrow 2x \ne k\pi \Leftrightarrow x \ne {{k\pi } \over 2}\,\,\left( {k \in Z} \right) \cr & PT \Leftrightarrow \tan x - {2 \over {\tan x}} + 1 = 0 \cr & \Leftrightarrow {\tan ^2}x + \tan x - 2 = 0 \cr & \Leftrightarrow \left[ \matrix{ \tan x = 1 \hfill \cr \tan x = - 2 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = {\pi \over 4} + k\pi \hfill \cr x = \arctan \left( { - 2} \right) + k\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} $$
