Đáp án:
$\left[\begin{array}{l}x = -\dfrac{\pi}{8}+ k\pi\\x =\dfrac{3\pi}{16} + k\dfrac{\pi}{2}\end{array}\right.\quad (k\in \Bbb Z)$
Giải thích các bước giải:
$\sin3x + \cos3x = \sqrt2\sin x$
$\Leftrightarrow \sin\left(3x + \dfrac{\pi}{4}\right) = \sin x$
$\Leftrightarrow \left[\begin{array}{l}3x + \dfrac{\pi}{4}= x + k2\pi\\3x + \dfrac{\pi}{4} = \pi - x + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = -\dfrac{\pi}{8}+ k\pi\\x =\dfrac{3\pi}{16} + k\dfrac{\pi}{2}\end{array}\right.\quad (k\in \Bbb Z)$
