
Giải chi tiết:
\(\begin{array}{l}
sinx - \sqrt 3 cosx = 2\left( {2co{s^2}x - 1} \right)\\
\Leftrightarrow sinx - \sqrt 3 cosx = 2cos2x\\
\Leftrightarrow \frac{{\sqrt 3 }}{2}cosx - \frac{1}{2}sinx = - cos2x\\
\Leftrightarrow \cos \left( {x + \frac{\pi }{6}} \right) = cos\left( {\pi - 2x} \right)\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{x + \frac{\pi }{6} = \pi - 2x + k2\pi }\\
{x + \frac{\pi }{6} = - \pi + 2x + k2\pi }
\end{array}} \right.;\left( {k \in Z} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{5\pi }}{{18}} + \frac{{k2\pi }}{3}\\
x = \frac{{7\pi }}{6} - k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)