A. \(x = \dfrac{{ - 5\pi }}{6} + k\pi ,\,x = \dfrac{{5\pi }}{6} + k2\pi ;\,k \in \mathbb{Z}\)
B. \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}x = k\pi ; \pm \dfrac{{5\pi }}{{12}} + k\pi ;k \in \mathbb{Z}\)
C. \(x = k\pi ;\,x = \dfrac{{5\pi }}{{12}} + k\pi ;\,k \in \mathbb{Z}\)
D. \(x = \dfrac{{ - 5\pi }}{{12}} + k\dfrac{\pi }{2};k \in \mathbb{Z}\)
B
Ta có: \(2{\cos ^2}2x + \left( {\sqrt 3 - 2} \right)\cos 2x - \sqrt 3 = 0\)
\( \Leftrightarrow \left( {\cos 2x - 1} \right)\left( {2\cos x + \sqrt 3 } \right) = 0\)
\(\Leftrightarrow \left[ \begin{array}{l}\cos 2x = 1\\\cos 2x = - \dfrac{{\sqrt 3 }}{2}\end{array} \right. \)
\( \Leftrightarrow \left[ \begin{array}{l} 2x = k2\pi \\ 2x = \pm \frac{{5\pi }}{6} + k2\pi \end{array} \right.\)
\(\Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x = \pm \dfrac{{5\pi }}{{12}} + k\pi \end{array} \right.\;\left( {k \in \mathbb{Z}} \right)\)
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