$\begin{array}{l}
\sqrt 3 \sin x - \cos x = 1\\
\Leftrightarrow 2\left( {\dfrac{{\sqrt 3 }}{2}\sin x - \dfrac{1}{2}\cos x} \right) = 1\\
\Leftrightarrow 2\sin \left( {x - \dfrac{\pi }{6}} \right) = 1\\
\Leftrightarrow \sin \left( {x - \dfrac{\pi }{6}} \right) = \dfrac{1}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{\pi }{6} = \dfrac{\pi }{6} + k2\pi \\
x - \dfrac{\pi }{6} = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k2\pi \\
x = \pi + k2\pi
\end{array} \right.\left( {k \in \mathbb{Z}} \right)
\end{array}$
