#andy
\[\begin{array}{l}
\sqrt 3 \sin x - \cos x + 1 = 0\\
\Leftrightarrow \frac{{\sqrt 3 }}{2}\sin x - \frac{1}{2}\cos x = - \frac{1}{2}\\
\Leftrightarrow \sin \left( {x - \frac{\pi }{6}} \right) = - \frac{1}{2}\\
\Leftrightarrow \{ \begin{array}{*{20}{c}}
{x - \frac{\pi }{6} = - \frac{\pi }{6} + k2\pi }\\
{x - \frac{\pi }{6} = \frac{{7\pi }}{6} + k2\pi }
\end{array}\\
\Leftrightarrow \{ \begin{array}{*{20}{c}}
{x = k2\pi }\\
{x = \frac{4}{3}\pi + k2\pi }
\end{array}
\end{array}\]
