Chủ đề này có 2 trả lời

[nguyen anh mai]

$2\sqrt{2}\left ( 1-cosxcos3x \right )= \frac{cos4x}{2cos^{2}\left ( x-\frac{3\pi }{8} \right )-1}$

[mystery266]

$2\sqrt{2}\left ( 1-cosxcos3x \right )= \frac{cos4x}{2cos^{2}\left ( x-\frac{3\pi }{8} \right )-1}$

$PT\Leftrightarrow \sqrt{2}\left ( 2-cos4x-cos2x \right )= \frac{cos4x}{cos( 2x-\frac{3\pi }{4})}$

 

$\Leftrightarrow  2-cos4x-cos2x = \frac{(cos2x-sin2x)(cos2x+sin2x)}{sin2x-cos2x}$

 

$\Leftrightarrow 2-cos4x-cos2x= -cos2x-sin2x$

 

$\Leftrightarrow 1+2sin^{2}2x= -sin2x$

 

ngon rồi

[Ha Manh Huu]

Thiếu ĐK: $cos\left ( 2x-\frac{3\pi }{4} \right )\neq 0$