giải pt: tan x – 2cot x +1 = 0 giúp vs ạ

giải pt: tan x – 2cot x +1 = 0 giúp vs ạ

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  1. \[\begin{array}{l}
    \tan x – 2\cot x + 1 = 0\\
    DK:\,\,\,\left\{ \begin{array}{l}
    \cos x \ne 0\\
    \sin x \ne 0
    \end{array} \right. \Leftrightarrow \sin 2x \ne 0.\\
    \Leftrightarrow \tan x – 2.\frac{1}{{\tan x}} + 1 = 0\\
    \Leftrightarrow {\tan ^2}x + \tan x – 2 = 0\\
    \Leftrightarrow \left( {\tan x – 1} \right)\left( {\tan x + 2} \right) = 0\\
    \Leftrightarrow \left[ \begin{array}{l}
    \tan x – 1 = 0\\
    \tan x + 2 = 0
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    \tan x = 1\\
    \tan x = – 2
    \end{array} \right..
    \end{array}\]

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  2. $$\eqalign{
    & \tan x – \cot x + 1 = 0\, \cr
    & DKXD:\,\left\{ \matrix{
    \sin x \ne 0 \hfill \cr
    \cos x \ne 0 \hfill \cr} \right. \Leftrightarrow \sin 2x \ne 0 \cr
    & \Leftrightarrow 2x \ne k\pi \Leftrightarrow x \ne {{k\pi } \over 2}\,\,\left( {k \in Z} \right) \cr
    & PT \Leftrightarrow \tan x – {2 \over {\tan x}} + 1 = 0 \cr
    & \Leftrightarrow {\tan ^2}x + \tan x – 2 = 0 \cr
    & \Leftrightarrow \left[ \matrix{
    \tan x = 1 \hfill \cr
    \tan x = – 2 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
    x = {\pi \over 4} + k\pi \hfill \cr
    x = \arctan \left( { – 2} \right) + k\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} $$

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