Nghiệm của phương trình \(2\cos 2x - 1 = 0\) là:

Ta có: \[{\sin ^2}\alpha  + {\cos ^2}\alpha  = 1 \Rightarrow \cos \alpha  =  \pm \frac{3}{5}\]. Vì \[ - \frac{\pi }{2} \le \alpha  \le 0 \Rightarrow \cos \alpha  > 0\].Suy ra \[\cos \alpha  = \frac{3}{5}\]

\[\cos 2\alpha  =  - \frac{7}{{25}}\]; \[\sin 2\alpha  =  - \frac{{24}}{{25}}\]

\[T =  - \frac{7}{6}\].

a) \[2\sin 2x + 1 = 0 \Leftrightarrow \sin 2x =  - \frac{1}{2} \Leftrightarrow \left\{ \begin{array}{l}2x =  - \frac{\pi }{6} + k2\pi \\2x = \pi  + \frac{\pi }{6} + k2\pi \end{array} \right.\]

Giải được \[x =  - \frac{\pi }{{12}} + k\pi ;\,x = \frac{{7\pi }}{{12}} + k\pi \,\,\,\left( {k \in \mathbb{Z}} \right)\]. Kết luận.

b) \[\cos \left( {2x - \frac{\pi }{6}} \right) = \cos \left( {x + \frac{\pi }{4}} \right) \Leftrightarrow \left\{ \begin{array}{l}2x - \frac{\pi }{6} = x + \frac{\pi }{4} + k2\pi \\2x - \frac{\pi }{6} =  - x - \frac{\pi }{4} + k2\pi \end{array} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}x = \frac{{5\pi }}{{12}} + k2\pi \\x =  - \frac{\pi }{{36}} + \frac{{k2\pi }}{3}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\] . Kết luận.

c) \[\cos x - \sqrt 3 \sin x =  - \sqrt 2  \Leftrightarrow \frac{1}{2}\cos x - \frac{{\sqrt 3 }}{2}\sin x =  - \frac{{\sqrt 2 }}{2} \Leftrightarrow \sin \left( {\frac{\pi }{6} - x} \right) = \sin \left( { - \frac{\pi }{4}} \right)\]

\[ \Leftrightarrow \left\{ \begin{array}{l}\frac{\pi }{6} - x =  - \frac{\pi }{4} + k2\pi \\\frac{\pi }{6} - x = \pi  + \frac{\pi }{4} + k2\pi \end{array} \right.\,\, \Leftrightarrow \left\{ \begin{array}{l}x = \frac{{5\pi }}{{12}} - k2\pi \\x =  - \frac{{13\pi }}{{12}} - k2\pi \end{array} \right.\,\left( {k \in \mathbb{Z}} \right)\]. Kết luận.