Đáp án:
\[\left[ \begin{array}{l}
x = \dfrac{{7\pi }}{{12}} + k\pi \\
x = - \dfrac{\pi }{4} + k\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
\(\begin{array}{l}
\cos 2x + \sqrt 3 \sin 2x = - \sqrt 3 \\
\Leftrightarrow \dfrac{1}{2}\cos 2x + \dfrac{{\sqrt 3 }}{2}\sin 2x = - \dfrac{{\sqrt 3 }}{2}\\
\Leftrightarrow \cos 2x.\cos \dfrac{\pi }{3} + \sin 2x.\sin \dfrac{\pi }{3} = - \dfrac{{\sqrt 3 }}{2}\\
\Leftrightarrow \cos \left( {2x - \dfrac{\pi }{3}} \right) = \cos \dfrac{{5\pi }}{6}\\
\Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{\pi }{3} = \dfrac{{5\pi }}{6} + k2\pi \\
2x - \dfrac{\pi }{3} = - \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{{7\pi }}{6} + k2\pi \\
2x = - \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{7\pi }}{{12}} + k\pi \\
x = - \dfrac{\pi }{4} + k\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
