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Điều kiện $\cos x\ne 0, \sin x \ne 0,1$.
PT $\Leftrightarrow \frac{\sin ^4 \frac{ x}{ 2} + \cos ^4 \frac{ x}{ 2} }{1 - \sin x } - \tan ^2 x . \sin x = \frac{1+\sin x }{ 2} + \tan ^2 x $
$\Leftrightarrow \frac{1-2\sin ^2 \frac{ x}{ 2} \cos ^2 \frac{ x}{ 2} }{1 - \sin x } = \frac{1+\sin x }{ 2} + \tan ^2 x (1+\sin x) $
$\Leftrightarrow \frac{1-\frac{1}{2}\sin ^2x}{1 - \sin x } =\frac{1}{2}(1+ \tan ^2 x) (1+\sin x)$
$\Leftrightarrow 1-\frac{1}{2}\sin ^2x=\frac{1}{2}(1+ \tan ^2 x) (1-\sin^2 x)$
$\Leftrightarrow 1-\frac{1}{2}\sin ^2x=\frac{1}{2}(1+ \tan ^2 x) \cos^2 x$
$\Leftrightarrow 1-\frac{1}{2}\sin ^2x=\frac{1}{2}(1+ \tan ^2 x) \cos^2 x$
$\Leftrightarrow2 \sin^2 x+ \cos^2 x=2$
$\Leftrightarrow \sin^2 x=1$
$\Leftrightarrow \sin x=-1$
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