Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{10} + k\dfrac{2\pi}{5}\\x = \dfrac{\pi}{2} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$\sin3x - \cos2x = 0$
$\Leftrightarrow \sin3x = \cos2x$
$\Leftrightarrow \sin3x = \sin\left(\dfrac{\pi}{2} - 2x\right)$
$\Leftrightarrow \left[\begin{array}{l}3x = \dfrac{\pi}{2} - 2x + k2\pi\\3x = \pi - \dfrac{\pi}{2} + 2x + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{10} + k\dfrac{2\pi}{5}\\x = \dfrac{\pi}{2} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)$
