Đáp án:
$\begin{array}{l}
\cos x - \sqrt 3 \sin x = - 1\\
\Leftrightarrow \dfrac{1}{2}\cos x - \dfrac{{\sqrt 3 }}{2}\sin x = - \dfrac{1}{2}\\
\Leftrightarrow \sin \dfrac{\pi }{6}.\cos x - \cos \dfrac{\pi }{6}.\sin x = - \dfrac{1}{2}\\
\Leftrightarrow \sin \left( {\dfrac{\pi }{6} - x} \right) = \sin - \dfrac{\pi }{6}\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{\pi }{6} - x = \dfrac{{ - \pi }}{6} + k2\pi \\
\dfrac{\pi }{6} - x = \pi + \dfrac{\pi }{6} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} - k2\pi \\
x = - \pi - k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
Vậy\,\left[ \begin{array}{l}
x = \dfrac{\pi }{3} - k2\pi \\
x = - \pi - k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}$
