Đáp án:
\(\left[ \begin{array}{l}
x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{7\pi }}{{54}} + \dfrac{{k2\pi }}{8}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\sqrt 3 \sin 6x - \cos 6x = 2\sin 3x\\
\to \dfrac{{\sqrt 3 }}{2}\sin 6x - \dfrac{1}{2}\cos 6x = \sin 3x\\
\to \sin 6x.\cos \dfrac{\pi }{6} - \cos 6x.\sin \dfrac{\pi }{6} = \sin 3x\\
\to \sin \left( {6x - \dfrac{\pi }{6}} \right) = \sin 3x\\
\to \left[ \begin{array}{l}
6x - \dfrac{\pi }{6} = 3x + k2\pi \\
6x - \dfrac{\pi }{6} = \pi - 3x + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
3x = \dfrac{\pi }{6} + k2\pi \\
9x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{7\pi }}{{54}} + \dfrac{{k2\pi }}{8}
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
