Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
f\left( x \right) = 3\cos 2x + \cos x + 1\\
= 3\left( {2{{\cos }^2}x - 1} \right) + \cos x + 1\\
= 6{\cos ^2}x + \cos x - 2\\
= 6.{\left( {\cos x + \frac{1}{{12}}} \right)^2} - \frac{{49}}{{24}} \ge - \frac{{49}}{{24}}\\
\Rightarrow f{\left( x \right)_{\min }} = - \frac{{49}}{{24}} \Leftrightarrow \cos x = - \frac{1}{{12}}\\
f\left( x \right) = 6{\cos ^2}x + \cos x - 2\\
\cos x \le 1 \Rightarrow 6{\cos ^2}x + \cos x - 2 \le {6.1^2} + 1 - 2 = 5\\
\Rightarrow f{\left( x \right)_{\max }} = 5 \Leftrightarrow \cos x = 1
\end{array}\)
