Đáp án:
\(\left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \dfrac{{7\pi }}{{12}} + k\pi
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\cos \left( {2x - \dfrac{{3\pi }}{2}} \right) + \sqrt 3 \cos 2x + 1 = 0\\
\to \cos \left( {2x - 2\pi + \dfrac{\pi }{2}} \right) + \sqrt 3 \cos 2x + 1 = 0\\
\to \cos \left( {2x + \dfrac{\pi }{2}} \right) + \sqrt 3 \cos 2x + 1 = 0\\
\to - \sin 2x + \sqrt 3 \cos 2x + 1 = 0\\
\to - \dfrac{1}{2}\sin 2x + \dfrac{{\sqrt 3 }}{2}\cos 2x + \dfrac{1}{2} = 0\\
\to - \sin 2x.\cos \dfrac{\pi }{3} + \cos 2x.\sin \dfrac{\pi }{3} = - \dfrac{1}{2}\\
\to \sin 2x.\cos \dfrac{\pi }{3} - \cos 2x.\sin \dfrac{\pi }{3} = \dfrac{1}{2}\\
\to \sin \left( {2x - \dfrac{\pi }{3}} \right) = \dfrac{1}{2}\\
\to \left[ \begin{array}{l}
2x - \dfrac{\pi }{3} = \dfrac{\pi }{6} + k2\pi \\
2x - \dfrac{\pi }{3} = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \dfrac{{7\pi }}{{12}} + k\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
