PT $\Leftrightarrow 2\sin x +2\sin x \cos 2x =(1-\sin 2x) + 2\cos x$
$\Leftrightarrow 2(\cos x -\sin x )+(\cos x - \sin x)^2 -2\sin x (\cos^2 x -\sin^2 x)=0$
$\Leftrightarrow (\cos x -\sin x) \bigg [ 2+\cos x -\sin x -2sin x(\cos x + \sin x) \bigg ]=0$
$\Leftrightarrow (\cos x -\sin x) \bigg [ \cos x -\sin x -2\sin x \cos x +2\cos^2 x \bigg ]=0$
Trong đó $\cos x -\sin x -2\sin x \cos x +2\cos^2 x=0$
$\Leftrightarrow \cos x -\sin x -\sin 2x + 1 +\cos 2x=0$
$\Leftrightarrow (\cos x -\sin x ) +(\cos^2 x -\sin^2 x) + (\cos x -\sin x)^2=0$
$\Leftrightarrow (\cos x-\sin x )(1 +\cos x +\sin x + \cos x -\sin x)=0$
$\Leftrightarrow (\cos x -\sin x)(1+2\cos x )=0$
Dễ rồi nhé