Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{4} + k\pi\\x = \dfrac{7\pi}{36} + k\dfrac{\pi}{3}\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\sin4x-\sqrt3\cos4x = \cos2x +\sqrt3\sin2x$
$\to \dfrac12\sin4x -\dfrac{\sqrt3}{2}\cos4x = \dfrac12\cos2x +\dfrac{\sqrt3}{2}\sin2x$
$\to \sin\left(4x -\dfrac{\pi}{3}\right) = \sin\left(2x + \dfrac{\pi}{6}\right)$
$\to \left[\begin{array}{l}4x -\dfrac{\pi}{3} = 2x +\dfrac{\pi}{6} + k2\pi\\4x -\dfrac{\pi}{3} = \dfrac{5\pi}{6} - 2x + k2\pi\end{array}\right.$
$\to \left[\begin{array}{l}2x = \dfrac{\pi}{2} + k2\pi\\6x = \dfrac{7\pi}{6} + k2\pi\end{array}\right.$
$\to \left[\begin{array}{l}x = \dfrac{\pi}{4} + k\pi\\x = \dfrac{7\pi}{36} + k\dfrac{\pi}{3}\end{array}\right.\quad (k\in\Bbb Z)$
