Đáp án:
\[\left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k\pi \\
x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
y = \dfrac{{\sin 3x}}{3} + \cos x - \sqrt 3 .\left( {\sin x + \dfrac{{\cos 3x}}{3}} \right)\\
\Rightarrow y' = \dfrac{{\left( {3x} \right)'.\cos 3x}}{3} - \sin x - \sqrt 3 .\left( {\cos x + \dfrac{{\left( {3x} \right)'.\left( { - \sin 3x} \right)}}{3}} \right)\\
\Leftrightarrow y' = \cos 3x - \sin x - \sqrt 3 \left( {\cos x - \sin 3x} \right)\\
y' = 0 \Leftrightarrow \cos 3x - \sin x - \sqrt 3 \cos x + \sqrt 3 \sin 3x = 0\\
\Leftrightarrow \cos 3x + \sqrt 3 \sin 3x = \sin x + \sqrt 3 \cos x\\
\Leftrightarrow \dfrac{1}{2}\cos 3x + \dfrac{{\sqrt 3 }}{2}\sin 3x = \dfrac{1}{2}\sin x + \dfrac{{\sqrt 3 }}{2}\cos x\\
\Leftrightarrow \sin \dfrac{\pi }{6}.\cos 3x + \sin 3x.\cos \dfrac{\pi }{6} = \sin x.\cos \dfrac{\pi }{3} + \cos x.\sin \dfrac{\pi }{3}\\
\Leftrightarrow \sin \left( {3x + \dfrac{\pi }{6}} \right) = \sin \left( {x + \dfrac{\pi }{3}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
3x + \dfrac{\pi }{6} = x + \dfrac{\pi }{3} + k2\pi \\
3x + \dfrac{\pi }{6} = \pi - x - \dfrac{\pi }{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k\pi \\
x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}
\end{array} \right.
\end{array}\)
Vậy nghiệm của phương trình \(y' = 0\) là: \(\left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k\pi \\
x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}
\end{array} \right.\)
