Giải thích các bước giải:
\(\begin{array}{l}
a,\\
2.\cos 3x + 1 = 0\\
\Leftrightarrow \cos 3x = - \frac{1}{2}\\
\Leftrightarrow 3x = \pm \frac{{2\pi }}{3} + k2\pi \\
\Leftrightarrow x = \pm \frac{{2\pi }}{9} + \frac{{k2\pi }}{3}\\
b,\\
\cos 2x - 5\cos x + 4 = 0\\
\Leftrightarrow \left( {2{{\cos }^2}x - 1} \right) - 5\cos x + 4 = 0\\
\Leftrightarrow 2{\cos ^2}x - 5\cos x + 3 = 0\\
\Leftrightarrow \left( {2\cos x - 3} \right)\left( {\cos x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = \frac{3}{2}\left( L \right)\\
\cos x = 1
\end{array} \right. \Leftrightarrow x = k2\pi \\
c,\\
\sqrt 3 \sin 2x + \cos 2x = - \sqrt 2 \\
\Leftrightarrow \frac{{\sqrt 3 }}{2}\sin 2x + \frac{1}{2}\cos 2x = - \frac{{\sqrt 2 }}{2}\\
\Leftrightarrow \sin \left( {2x + \frac{\pi }{6}} \right) = \sin \left( { - \frac{\pi }{4}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
2x + \frac{\pi }{6} = - \frac{\pi }{4} + k2\pi \\
2x + \frac{\pi }{6} = \frac{{5\pi }}{4} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - \frac{5}{{24}}\pi + k\pi \\
x = \frac{{13}}{{24}}\pi + k\pi
\end{array} \right.
\end{array}\)
