Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\cos 2x + \cos x = 4{\sin ^2}\dfrac{x}{2} - 1\\
\Leftrightarrow \left( {2{{\cos }^2}x - 1} \right) + \cos x = 2\left( {2{{\sin }^2}\dfrac{x}{2} - 1} \right) + 1\\
\Leftrightarrow 2{\cos ^2}x - 1 + \cos x = - 2\cos x + 1\\
\Leftrightarrow 2{\cos ^2}x + 3\cos x - 2 = 0\\
\Leftrightarrow \left( {2\cos x - 1} \right)\left( {\cos x + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = \dfrac{1}{2}\\
\cos x = - 2\,\,\,\,\,\,\left( {L,\,\,\, - 1 \le \cos x \le 1} \right)
\end{array} \right.\\
\Leftrightarrow \cos x = \dfrac{1}{2}\\
\Leftrightarrow x = \pm \dfrac{\pi }{3} + k2\pi \\
b,\\
2\cos 2x.\cos x = 1 + \cos 2x + \cos 3x\\
\Leftrightarrow \cos \left( {2x + x} \right) + \cos \left( {2x - x} \right) = 1 + \cos 2x + \cos 3x\\
\Leftrightarrow \cos 3x + \cos x = 1 + \cos 2x + \cos 3x\\
\Leftrightarrow \cos x = 1 + \cos 2x\\
\Leftrightarrow \cos x = 1 + \left( {2{{\cos }^2}x - 1} \right)\\
\Leftrightarrow 2{\cos ^2}x - \cos x = 0\\
\Leftrightarrow \cos x\left( {2\cos x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\cos x = \dfrac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \pm \dfrac{\pi }{3} + k2\pi
\end{array} \right.
\end{array}\)
