Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{4} + k2\pi\\x =\dfrac{\pi}{20} + k\dfrac{2\pi}{5}\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\quad -\cos3x -\sin3x +\sqrt2\cos2x = 0$
$\to \dfrac{\sqrt2}{2}\cos3x +\dfrac{\sqrt2}{2}\sin3x = \cos2x$
$\to \cos\left(3x -\dfrac{\pi}{4}\right) = \cos2x$
$\to \left[\begin{array}{l}3x -\dfrac{\pi}{4} = 2x + k2\pi\\3x -\dfrac{\pi}{4} = -2x + k2\pi\end{array}\right.$
$\to \left[\begin{array}{l}x = \dfrac{\pi}{4} + k2\pi\\x =\dfrac{\pi}{20} + k\dfrac{2\pi}{5}\end{array}\right.\quad (k\in\Bbb Z)$
