cos2x=sin4x có nghiệm

Đáp án:

$\left[\begin{array}{l}x = \dfrac{\pi}{12}+ k\dfrac{\pi}{3}\\x = \dfrac{\pi}{4} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$

Giải thích các bước giải:

$\cos2x = \sin4x$

$\Leftrightarrow \cos2x = \cos\left(\dfrac{\pi}{2} - 4\right)$

$\Leftrightarrow \left[\begin{array}{l}2x = \dfrac{\pi}{2} - 4x + k2\pi\\2x = 4x - \dfrac{\pi}{2} + k2\pi\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}6x = \dfrac{\pi}{2}+ k2\pi\\2x = \dfrac{\pi}{2} + k2\pi\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{12}+ k\dfrac{\pi}{3}\\x = \dfrac{\pi}{4} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$