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