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Điều kiện : $\cos x,\cos (\frac{\pi}{2}+x ) \ne 0$ PT $\Leftrightarrow -\tan (\pi -\frac{\pi}{2}-x )-3 \tan^2x=\frac{-2\sin^2 x }{\cos^2x} $ $\Leftrightarrow- \tan (\frac{\pi}{2}-x )-3 \tan^2x=-2 \tan^2x$ $\Leftrightarrow -\cot x= \tan^2x$ $\Leftrightarrow -1= \tan^3x$ $\Leftrightarrow -1= \tan x$ $\Leftrightarrow x=\frac{-\pi }{4}+ k\pi ( k\in \mathbb{Z}).$
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