Đáp án:

\(\left\{ \begin{array}{l}
x = \dfrac{\pi }{6} + k\pi \\
x = \dfrac{\pi }{3} + k2\pi \\
x = \pi  + k2\pi 
\end{array} \right.\) $(k\in\mathbb Z)$

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
2{\sin ^2}x - \sqrt 3 \sin 2x + 1 = \sqrt 3 \sin x - \cos x\\
 \Leftrightarrow 2{\sin ^2}x - 2\sqrt 3 \sin x\cos x + \left( {{{\sin }^2}x + {{\cos }^2}x} \right) = \sqrt 3 \sin x - \cos x\\
 \Leftrightarrow 3{\sin ^2}x - 2\sqrt 3 \sin x\cos x + {\cos ^2}x = \sqrt 3 \sin x - \cos x\\
 \Leftrightarrow {\left( {\sqrt 3 \sin x - \cos x} \right)^2} = \sqrt 3 \sin x - \cos x\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt 3 \sin x - \cos x = 0\\
\sqrt 3 \sin x - \cos x = 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{{\sqrt 3 }}{2}\sin x - \dfrac{1}{2}\cos x = 0\\
\dfrac{{\sqrt 3 }}{2}\sin x - \dfrac{1}{2}\cos x = \dfrac{1}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin \left( {x - \dfrac{\pi }{6}} \right) = 0\\
\sin \left( {x - \dfrac{\pi }{6}} \right) = \sin \dfrac{\pi }{6}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{\pi }{6} = k\pi \\
x - \dfrac{\pi }{6} = \dfrac{\pi }{6} + k2\pi \\
x - \dfrac{\pi }{6} = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k\pi \\
x = \dfrac{\pi }{3} + k2\pi \\
x = \pi  + k2\pi 
\end{array} \right.
\end{array}\) $(k\in\mathbb Z)$.