Đáp án:
\[\left[ \begin{array}{l}
x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{9}\\
x = \dfrac{{7\pi }}{{54}} + \dfrac{{k2\pi }}{9}
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sin 3x = 3\sin x - 4{\sin ^3}x\\
3\sin 3x - \sqrt 3 \cos 9x = 1 + 4{\sin ^3}3x\\
\sin \left( {x - y} \right) = \sin x.\cos y - \sin y.\cos x\\
\Leftrightarrow \left( {3\sin 3x - 4{{\sin }^3}3x} \right) - \sqrt 3 \cos 9x = 1\\
\Leftrightarrow \sin 9x - \sqrt 3 \cos 9x = 1\\
\Leftrightarrow \dfrac{1}{2}\sin 9x - \dfrac{{\sqrt 3 }}{2}\cos 9x = \dfrac{1}{2}\\
\Leftrightarrow \sin 9x.\cos \dfrac{\pi }{3} - \cos 9x.\sin \dfrac{\pi }{3} = \sin \dfrac{\pi }{6}\\
\Leftrightarrow \sin \left( {9x - \dfrac{\pi }{3}} \right) = \sin \dfrac{\pi }{6}\\
\Leftrightarrow \left[ \begin{array}{l}
9x - \dfrac{\pi }{3} = \dfrac{\pi }{6} + k2\pi \\
9x - \dfrac{\pi }{3} = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{9}\\
x = \dfrac{{7\pi }}{{54}} + \dfrac{{k2\pi }}{9}
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
