Đáp án:
$\left[\begin{array}{l}x =-\dfrac{\pi}{2} + k2\pi\\x = \dfrac{3\pi}{10} + k\dfrac{2\pi}{5}\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\sin3x =-\cos2x$
$\Leftrightarrow \sin3x =\cos(\pi - 2x)$
$\Leftrightarrow \sin3x =\sin\left(-\dfrac{\pi}{2} + 2x\right)$
$\Leftrightarrow \left[\begin{array}{l}3x =-\dfrac{\pi}{2} + 2x + k2\pi\\3x = \dfrac{3\pi}{2} - 2x + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x =-\dfrac{\pi}{2} + k2\pi\\x = \dfrac{3\pi}{10} + k\dfrac{2\pi}{5}\end{array}\right.\quad (k\in\Bbb Z)$
