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Điều kiện $\cos x \ne 0.$
PT
$\Leftrightarrow (\sin x -\cos x)^2 =2\sin^{2}x -\dfrac{\sin x}{\cos x}$
$\Leftrightarrow \sin^2 x +\cos^2 x-2\sin x \cos x =\dfrac{\sin x}{\cos x}(2\sin x \cos x -1) $
$\Leftrightarrow 1-\sin 2x +\dfrac{\sin x}{\cos x}(1-\sin 2x )=0 $
$\Leftrightarrow (1+\tan x)(1-\sin 2x )=0 $
$\Leftrightarrow \left[ {\begin{matrix} \tan x=-1\\ \sin 2x =1 \end{matrix}} \right.$
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