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Điều kiện: $\left\{ \begin{array}{l} \sin 2x\ne0\\ \tan x\ne-1 \end{array} \right.$
Ta có:
$\cot x-1=\frac{\cos2x}{1+\tan x}+\sin^{2}x-\frac{1}{2}\sin2x$
$\Leftrightarrow \cot x-1=\frac{\cos^2x-\sin^2x}{\displaystyle 1+\frac{\sin x}{\cos x}}+\sin^{2}x-\frac{1}{2}\sin2x$
$\Leftrightarrow \cot x-1=\cos x(\cos x-\sin x)+\sin^2x-\frac{1}{2}\sin2x$
$\Leftrightarrow \cot x-1=1-\sin2x$
$\Leftrightarrow \frac{1}{\tan x}=2-\frac{2\tan x}{1+\tan^2x}$
$\Leftrightarrow 2\tan^3x-3\tan^2x+2\tan x-1=0$
$\Leftrightarrow \tan x=1\Leftrightarrow x=\frac{\pi}{4}+k\pi,k\in\mathbb{Z}$
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