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