Đáp án:
$\left[\begin{array}{l}x = k2\pi\\x = \dfrac{4\pi}{3} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\quad \sqrt3\sin x -\cos x = -1$
$\to \dfrac{\sqrt3}{2}\sin x -\dfrac12\cos x = -\dfrac12$
$\to \sin\left(x -\dfrac{\pi}{6}\right) = \sin\left(-\dfrac{\pi}{6}\right)$
$\to \left[\begin{array}{l}x -\dfrac{\pi}{6} =-\dfrac{\pi}{6} + k2\pi\\x -\dfrac{\pi}{6} = \dfrac{7\pi}{6} + k2\pi\end{array}\right.$
$\to \left[\begin{array}{l}x = k2\pi\\x = \dfrac{4\pi}{3} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
