Đáp án:
$\left[\begin{array}{l}x =\dfrac{\pi}{6} + k\dfrac{2\pi}{3}\\x= -\dfrac{\pi}{42}+ k\dfrac{2\pi}{7}\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\sin5x +\sqrt3\cos5x =\cos2x -\sqrt3\sin2x$
$\to \dfrac12\sin5x +\dfrac{\sqrt3}{2}\cos5x =\dfrac{1}{2}\cos2x -\dfrac{\sqrt3}{2}\sin2x$
$\to \cos\left(5x -\dfrac{\pi}{6}\right) =\cos\left(2x +\dfrac{\pi}{3}\right)$
$\to \left[\begin{array}{l}5x -\dfrac{\pi}{6}=2x +\dfrac{\pi}{3} + k2\pi\\5x -\dfrac{\pi}{6}=- 2x -\dfrac{\pi}{3} + k2\pi\end{array}\right.$
$\to \left[\begin{array}{l}3x =\dfrac{\pi}{2} + k2\pi\\7x= -\dfrac{\pi}{6}+ k2\pi\end{array}\right.$
$\to \left[\begin{array}{l}x =\dfrac{\pi}{6} + k\dfrac{2\pi}{3}\\x= -\dfrac{\pi}{42}+ k\dfrac{2\pi}{7}\end{array}\right.\quad (k\in\Bbb Z)$
