Đáp án:
$\left[ \begin{array}{l}
x = \frac{{5\pi }}{{12}} + k2\pi \\
x = \frac{{11\pi }}{{12}} + k2\pi
\end{array} \right.\,\,\,\,\,\left( {k \in } \mathbb{Z} \right)$
Giải thích các bước giải:
\(\begin{array}{l}\sqrt 3 \sin x - \cos x - \sqrt 2 = 0\\ \Leftrightarrow \frac{{\sqrt 3 }}{2}\sin x - \frac{1}{2}\cos x = \frac{{\sqrt 2 }}{2}\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{6}} \right) = \frac{{\sqrt 2 }}{2}\\ \Leftrightarrow \left[ \begin{array}{l}x - \frac{\pi }{6} = \frac{\pi }{4} + k2\pi \\x - \frac{\pi }{6} = \frac{{3\pi }}{4} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{{5\pi }}{{12}} + k2\pi \\x = \frac{{11\pi }}{{12}} + k2\pi \end{array} \right.\,\,\,\,\,\left( {k \in \mathbb{Z}} \right).\end{array}\)
