Đáp án:
\(\left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k2\pi \\
x = \dfrac{{7\pi }}{{12}} + k2\pi
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\sqrt 3 \sin x + \cos x = \sqrt 2 \\
\to \dfrac{{\sqrt 3 }}{2}\sin x + \dfrac{1}{2}\cos x = \dfrac{{\sqrt 2 }}{2}\\
\to \sin x.\cos \dfrac{\pi }{6} + \sin \dfrac{\pi }{6}.\cos x = \dfrac{{\sqrt 2 }}{2}\\
\to \sin \left( {x + \dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 2 }}{2}\\
\to \left[ \begin{array}{l}
x + \dfrac{\pi }{6} = \dfrac{\pi }{4} + k2\pi \\
x + \dfrac{\pi }{6} = \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k2\pi \\
x = \dfrac{{7\pi }}{{12}} + k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
