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:
$\sin^4\left(x +\dfrac{\pi}{2}\right) - \sin^4x = \sin4x$
$\Leftrightarrow \cos^4x - \sin^4x = \sin4x$
$\Leftrightarrow (\cos^2x - \sin^2x)(\cos^2x +\sin^2x) = \sin4x$
$\Leftrightarrow \cos2x = \sin4x$
$\Leftrightarrow \cos2x = \cos\left(\dfrac{\pi}{2} - 4x\right)$
$\Leftrightarrow \left[\begin{array}{l}2x = \dfrac{\pi}{2} - 4x + k2\pi\\2x = 4x -\dfrac{\pi}{2} + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{12} + k\dfrac{\pi}{3}\\x =\dfrac{\pi}{4} + k\pi\end{array}\right.\quad (k \in \Bbb Z)$
