Đáp án:
$C.\,\sin\left(3x + \dfrac{\pi}{6}\right) = -\dfrac{1}{2}$
Giải thích các bước giải:
$\sqrt3\sin3x + \cos3x = -1$
$\Leftrightarrow \dfrac{\sqrt3}{2}\sin3x + \dfrac{1}{2}\cos3x = -\dfrac{1}{2}$
$\Leftrightarrow \sin3x\cos\dfrac{\pi}{6} + \cos3x\sin\dfrac{\pi}{6} = -\dfrac{1}{2}$
$\Leftrightarrow \sin\left(3x + \dfrac{\pi}{6}\right) = -\dfrac{1}{2}$