Đáp án:
\[\left[ \begin{array}{l}
x = \frac{\pi }{6} + k\pi \\
x = \frac{\pi }{3} + k2\pi \\
x = \pi + k2\pi
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\[\begin{array}{l}
1 - \cos 2x - \sqrt 3 .2.\sin x.\cos x + 1 = \sqrt 3 \sin x - \cos x\\
\Leftrightarrow \left( {\sqrt 3 \sin x - \cos x} \right) + \left( {\sqrt 3 \sin 2x + \cos 2x} \right) = 2\\
\Leftrightarrow \left( {\frac{{\sqrt 3 }}{2}\sin x - \cos x} \right) + \left( {\frac{{\sqrt 3 }}{2}\sin 2x + \frac{1}{2}\cos 2x} \right) = 1\\
\Leftrightarrow \sin \left( {x - \frac{\pi }{6}} \right) + \cos \left( {2x - \frac{\pi }{3}} \right) = 1\\
\Leftrightarrow \sin \left( {x - \frac{\pi }{6}} \right) + 1 - 2{\sin ^2}\left( {x - \frac{\pi }{6}} \right) = 1\\
\Leftrightarrow 2{\sin ^2}\left( {x - \frac{\pi }{6}} \right) - \sin \left( {x - \frac{\pi }{6}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin \left( {x - \frac{\pi }{6}} \right) = 0\\
\sin \left( {x - \frac{\pi }{6}} \right) = \frac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x - \frac{\pi }{6} = k\pi \\
x - \frac{\pi }{6} = \frac{\pi }{6} + k2\pi \\
x - \frac{\pi }{6} = \frac{{5\pi }}{6} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{6} + k\pi \\
x = \frac{\pi }{3} + k2\pi \\
x = \pi + k2\pi
\end{array} \right.
\end{array}\]
