(3 điểm) Giải các phương trình lượng giác cơ bản sau:

a) \(\cos x = \frac{1}{2}\);

b) \(\tan 2x = 1\);

c) \(2\left( {\sin x + 3} \right){\cos ^4}\frac{x}{2} - \sin x\left( {1 + \cos x} \right) - 3\cos x - 1 = 0\).

a) \(\cos x = \frac{1}{2}\)

\( \Leftrightarrow \cos x = \cos \frac{\pi }{3}\)

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

b) \(\tan 2x = 1\)

\( \Leftrightarrow \tan 2x = \tan \frac{\pi }{4}\)

\( \Leftrightarrow 2x = \frac{\pi }{4} + k\pi \,\,\left( {k \in \mathbb{Z}} \right)\)

\( \Leftrightarrow x = \frac{\pi }{8} + k\frac{\pi }{2}\,\,\left( {k \in \mathbb{Z}} \right)\).

c) \(2\left( {\sin x + 3} \right){\cos ^4}\frac{x}{2} - \sin x\left( {1 + \cos x} \right) - 3\cos x - 1 = 0\)

\( \Leftrightarrow 2\left( {\sin x + 3} \right){\cos ^4}\frac{x}{2} - 2\sin x{\cos ^2}\frac{x}{2} - 6{\cos ^2}\frac{x}{2} + 2 = 0\)

\( \Leftrightarrow \left( {\sin x + 3} \right){\cos ^4}\frac{x}{2} - {\cos ^2}\frac{x}{2}\left( {\sin x + 3} \right) + 1 = 0\)

\( \Leftrightarrow {\cos ^2}\frac{x}{2}\left( {\sin x + 3} \right)\left( {{{\cos }^2}\frac{x}{2} - 1} \right) + 1 = 0\)

\( \Leftrightarrow {\cos ^2}\frac{x}{2}\left( {\sin x + 3} \right)\left( { - {{\sin }^2}\frac{x}{2}} \right) + 1 = 0\)

\( \Leftrightarrow \left[ { - \frac{1}{4} \cdot {{\left( {2\sin \frac{x}{2}\cos \frac{x}{2}} \right)}^2}} \right]\left( {\sin x + 3} \right) + 1 = 0\)

\( \Leftrightarrow - \frac{1}{4}{\sin ^2}x\left( {\sin x + 3} \right) + 1 = 0\)

\( \Leftrightarrow {\sin ^3}x + 3{\sin ^2}x - 4 = 0\)

\( \Leftrightarrow \left( {\sin x - 1} \right){\left( {\sin x + 2} \right)^2} = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}\sin x = 1\\\sin x = - 2\,\,\left( {{\rm{vn}}} \right)\end{array} \right.\)

\( \Leftrightarrow x = \frac{\pi }{2} + k2\pi \,\,\left( {k \in \mathbb{Z}} \right)\).