Tan^2x+cot^2x+2(tanx+cotx)=6

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Đáp án:

$x = \arctan \left( { - 2 + \sqrt 3 } \right) + k\pi$;
$x = \arctan \left( { - 2 - \sqrt 3 } \right) + k\pi$ 
$\left( {k \in Z} \right)$

Lời giải:

$\eqalign{ & {\tan ^2}x + {\cot ^2}x + 2\left( {\tan x + \cot x} \right) = 6\, \cr & \text{Điều kiện xác định: }\left\{ \matrix{ \sin x \ne 0 \hfill \cr \cos x \ne 0 \hfill \cr} \right. \Leftrightarrow \sin 2x \ne 0 \Leftrightarrow x \ne {{k\pi } \over 2}\,\,\left( {k \in Z} \right) \cr & \text{Đặt }t = \tan x + \cot x \cr & \Rightarrow {t^2} = {\tan ^2}x + {\cot ^2}x + 2 \Rightarrow {\tan ^2}x + {\cot ^2}x = {t^2} - 2 \cr & \text{ Phương trình }{t^2} - 2 + 2t = 6 \Leftrightarrow {t^2} + 2t - 8 = 0 \Leftrightarrow \left[ \matrix{ t = 2 \hfill \cr t = - 4 \hfill \cr} \right. \cr & t = 2 \Leftrightarrow \tan x + \cot x = 2 \cr & \Leftrightarrow \tan x + {1 \over {\tan x}} = 2 \Leftrightarrow {\tan ^2}x - 2\tan x + 1 = 0 \Leftrightarrow \tan x = 1 \cr & \Leftrightarrow x = {\pi \over 4} + k\pi \,\,\left( {k \in Z} \right) \cr & t = - 4 \Leftrightarrow \tan x + \cot x = - 4 \cr & \Leftrightarrow {\mathop{\rm tanx}\nolimits} + {1 \over {\tan x}} = - 4 \Leftrightarrow {\tan ^2}x + 4\tan x + 1 = 0 \cr & \Leftrightarrow \left[ \matrix{ \tan x = - 2 + \sqrt 3 \hfill \cr \tan x = - 2 - \sqrt 3 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = \arctan \left( { - 2 + \sqrt 3 } \right) + k\pi \hfill \cr x = \arctan \left( { - 2 - \sqrt 3 } \right) + k\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr}$

Vậy $x = \arctan \left( { - 2 + \sqrt 3 } \right) + k\pi$;
$x = \arctan \left( { - 2 - \sqrt 3 } \right) + k\pi$ 
$\left( {k \in Z} \right)$.