Đáp án:
\[x = \dfrac{\pi }{{36}} + \dfrac{{k\pi }}{3}\,\,\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sqrt 6 \cos 3x + \sqrt 2 \sin 3x = \sqrt 8 \cos 3x\\
\Leftrightarrow \sqrt 2 .\sqrt 3 .\cos 3x + \sqrt 2 \sin 3x = 2\sqrt 2 .\cos 3x\\
\Leftrightarrow \sqrt 2 .\left( {\sqrt 3 \cos 3x + \sin 3x} \right) = 2\sqrt 2 .\cos 3x\\
\Leftrightarrow \sqrt 3 \cos 3x + \sin 3x = 2\cos 3x\\
\Leftrightarrow \dfrac{{\sqrt 3 }}{2}\cos 3x + \dfrac{1}{2}\sin 3x = \cos 3x\\
\Leftrightarrow \cos 3x.\cos \dfrac{\pi }{6} + \sin 3x.\sin \dfrac{\pi }{6} = \cos 3x\\
\Leftrightarrow \cos \left( {3x - \dfrac{\pi }{6}} \right) = \cos 3x\\
\Leftrightarrow \left[ \begin{array}{l}
3x - \dfrac{\pi }{6} = 3x + k2\pi \\
3x - \dfrac{\pi }{6} = - 3x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
- \dfrac{\pi }{6} = k2\pi \\
6x = \dfrac{\pi }{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow x = \dfrac{\pi }{{36}} + \dfrac{{k\pi }}{3}\,\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
