Đáp án:
\[\left[ \begin{array}{l}
x = \dfrac{\pi }{6} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
\(\begin{array}{l}
\sin 2x - \sqrt 3 \cos 2x = 2\sin \left( {\dfrac{\pi }{6} - x} \right)\\
\Leftrightarrow \dfrac{1}{2}\sin 2x - \dfrac{{\sqrt 3 }}{2}\cos 2x = \sin \left( {\dfrac{\pi }{6} - x} \right)\\
\Leftrightarrow \sin 2x.\cos \dfrac{\pi }{3} - \cos 2x.\sin \dfrac{\pi }{3} = \sin \left( {\dfrac{\pi }{6} - x} \right)\\
\Leftrightarrow \sin \left( {2x - \dfrac{\pi }{3}} \right) = \sin \left( {\dfrac{\pi }{6} - x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{\pi }{3} = \dfrac{\pi }{6} - x + k2\pi \\
2x - \dfrac{\pi }{3} = \pi - \left( {\dfrac{\pi }{6} - x} \right) + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{\pi }{3} = \dfrac{\pi }{6} - x + k2\pi \\
2x - \dfrac{\pi }{3} = \dfrac{{5\pi }}{6} + x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x + x = \dfrac{\pi }{6} + \dfrac{\pi }{3} + k2\pi \\
2x - x = \dfrac{{5\pi }}{6} + \dfrac{\pi }{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{\pi }{2} + k2\pi \\
x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
