Hi vọng ai đó có ý tưởng tốt hơn
$\begin{array}{l}
3\tan 2x + 2\cos 2x = \dfrac{3}{{\cos 2x}} + 2\tan \left( {x - \dfrac{\pi }{4}} \right)\\
\Leftrightarrow 3\tan 2x + 2\cos 2x = \dfrac{3}{{\cos 2x}} + 2.\dfrac{{1 - \tan x}}{{1 + \tan x}}\\
t = \tan x \Rightarrow \tan 2x = \dfrac{{2t}}{{1 - {t^2}}},\cos 2x = \dfrac{{1 - {t^2}}}{{1 + {t^2}}}\\
PT \Leftrightarrow \dfrac{{6t}}{{1 - {t^2}}} + \dfrac{{2 - 2{t^2}}}{{1 + {t^2}}} = \dfrac{{3 + 3{t^2}}}{{1 - {t^2}}} + \dfrac{{2 - 2t}}{{1 + t}}\\
\Leftrightarrow 6t\left( {1 + {t^2}} \right) + \left( {2 - 2{t^2}} \right)\left( {1 - {t^2}} \right) = \left( {3 + 3{t^2}} \right)\left( {1 + {t^2}} \right) + \left( {2 - 2t} \right)\left( {1 - t} \right)\left( {1 + {t^2}} \right)\\
\Leftrightarrow 6t + 6{t^3} + 2 - 2{t^2} - 2{t^2} + 2{t^4} = 3{t^4} + 6{t^2} + 3 + 2\left( {{t^2} - 2t + 1} \right)\left( {1 + {t^2}} \right)\\
\Leftrightarrow 2{t^4} + 6{t^3} - 4{t^2} + 6t + 2 = 3{t^4} + 6{t^2} + 3 + 2\left( {{t^4} + {t^2} - 2t - 2{t^3} + 1 + {t^2}} \right)\\
\Leftrightarrow 2{t^4} + 6{t^3} - 4{t^2} + 6t + 2 = 5{t^4} - 4{t^3} + 10{t^2} - 4t + 5\\
\Leftrightarrow 3{t^4} - 10{t^3} + 14{t^2} - 10t + 3 = 0\\
\Leftrightarrow 3{t^4} - 3{t^3} - 7{t^3} + 7{t^2} + 7{t^2} - 7t - 3t + 3 = 0\\
\Leftrightarrow \left( {t - 1} \right)\left( {3{t^3} - 7{t^2} + 7t - 3} \right) = 0\\
\Leftrightarrow \left( {t - 1} \right)\left[ {3{t^3} - 3{t^2} - 4{t^2} + 4t + 3t - 3} \right] = 0\\
\Leftrightarrow {\left( {t - 1} \right)^2}\left( {3{t^2} - 4t + 3} \right) = 0\\
\Leftrightarrow t = 1\\
\Leftrightarrow \tan x = 1 \Leftrightarrow x = \dfrac{\pi }{4} + k\pi \left( {k \in \mathbb{Z}} \right)
\end{array}$
