Đáp án:
$\begin{array}{l}
y = - \cos x + \sqrt 3 \sin x - x - 5\\
\Rightarrow y' = \sin x + \sqrt 3 \cos x - 1 = 0\\
\Rightarrow \sin x + \sqrt 3 \cos x = 1\\
\Rightarrow \dfrac{1}{2}.\sin x + \dfrac{{\sqrt 3 }}{2}\cos x = \dfrac{1}{2}\\
\Rightarrow \sin \dfrac{\pi }{6}.\sin x + \cos \dfrac{\pi }{6}.\cos x = \dfrac{1}{2}\\
\Rightarrow \sin \left( {x + \dfrac{\pi }{6}} \right) = \dfrac{1}{2} = \sin \dfrac{\pi }{6}\\
\Rightarrow \left[ \begin{array}{l}
x = k2\pi \\
x = \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.
\end{array}$
