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DK: \sin x\neq 1
\frac{( 1- \cos x)^2+ ( 1 \\cos x)^}{4\ (1- \sin x )}-\tan^2x\sin x=\frac{1+\sin x}{2}+\tan^2 x
\Rightarrow \frac{1+\cos^2 x}{2 ( 1-\sin x)}-\tan^2 x\sin x-\frac{1+\sin x}{2}-\tan^2 x=0
\Rightarrow \frac{1+\cos^2 x}{2 ( 1-\sin x)}-\tan^2 x( \sin x \1 )-\frac{1+\sin x}{2}=0
\Rightarrow \frac{1+\cos^2 x-1+\sin^2 x}{2( 1- \sin x )}-\tan^2 x ( \sin x+ \1 )=0
\Rightarrow \frac{1}{2\( 1-\sin x)}-\frac{\sin^2 x}{\cos^2 x}\ ( \sin x +\1)=0
\Rightarrow \frac{1}{2\( 1-\sin x)}-\frac{\sin^2 x}{1-\sin^2 x}\( \sin x\1 )=0
\Rightarrow \frac{1-2\sin^2 x}{2\ ( 1-\sin x)}=0
\Rightarrow \frac{\cos 2x}{2\(1-\sin x )}=0 \Rightarrow \cos 2x=0 (Vi \sin x\neq 1) \Rightarrow x=\frac{ \pi }{4}+k\pi (k\in\mathbb{Z})$
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