$2\sqrt{2}\left ( 1-cosxcos3x \right )= \frac{cos4x}{2cos^{2}\left ( x-\frac{3\pi }{8} \right )-1}$
$PT\Leftrightarrow \sqrt{2}\left ( 2-cos4x-cos2x \right )= \frac{cos4x}{cos( 2x-\frac{3\pi }{4})}$
$\Leftrightarrow 2-cos4x-cos2x = \frac{(cos2x-sin2x)(cos2x+sin2x)}{sin2x-cos2x}$
$\Leftrightarrow 2-cos4x-cos2x= -cos2x-sin2x$
$\Leftrightarrow 1+2sin^{2}2x= -sin2x$
ngon rồi
- Ha Manh Huu và nguyen anh mai thích