Đáp án:

$\begin{array}{l}
\max y = 2 \Leftrightarrow x = \pi  + k2\pi \left( {k \in Z} \right)\\
\min y = \dfrac{{ - 9}}{8} \Leftrightarrow \left[ \begin{array}{l}
x = \arccos \left( {\dfrac{1}{4}} \right) + k2\pi \\
x =  - \arccos \left( {\dfrac{1}{4}} \right) + k2\pi 
\end{array} \right.
\end{array}$

Giải thích các bước giải:

 +) Tìm $\min$:

Ta có:

$\begin{array}{l}
y = \cos 2x - \cos x\\
 = 2{\cos ^2}x - \cos x - 1\\
 = 2\left( {{{\cos }^2}x - 2.\cos x.\dfrac{1}{4} + \dfrac{1}{{16}}} \right) - \dfrac{9}{8}\\
 = 2{\left( {\cos x - \dfrac{1}{4}} \right)^2} - \dfrac{9}{8}
\end{array}$

Do ${\left( {\cos x - \dfrac{1}{4}} \right)^2} \ge 0,\forall x$

$\begin{array}{l}
 \Rightarrow y \ge \dfrac{{ - 9}}{8}\\
 \Rightarrow \min y = \dfrac{{ - 9}}{8}\cos x - \dfrac{1}{4} = 0
\end{array}$

Dấu bằng xảy ra:

$ \Leftrightarrow \cos x = \dfrac{1}{4} \Leftrightarrow \left[ \begin{array}{l}
x = \arccos \left( {\dfrac{1}{4}} \right) + k2\pi \\
x =  - \arccos \left( {\dfrac{1}{4}} \right) + k2\pi 
\end{array} \right.$

+) Tìm $\max$:

Ta có:

$\begin{array}{l}
y = \cos 2x - \cos x\\
 = 2{\cos ^2}x - \cos x - 1
\end{array}$

Mà lại có: 

$\begin{array}{l}
 - 1 \le \cos x \le 1\\
 \Rightarrow \left\{ \begin{array}{l}
{\cos ^2}x \le 1\\
 - \cos x \le 1
\end{array} \right. \Rightarrow 2{\cos ^2}x - \cos x - 1 \le 2.1 + 1 - 1 = 2\\
 \Rightarrow y \le 2\\
 \Rightarrow \max y = 2
\end{array}$

Dấu bằng xảy ra:

$ \Leftrightarrow \cos x =  - 1 \Leftrightarrow x = \pi  + k2\pi $

Vậy $\begin{array}{l}
\max y = 2 \Leftrightarrow x = \pi  + k2\pi \left( {k \in Z} \right)\\
\min y = \dfrac{{ - 9}}{8} \Leftrightarrow \left[ \begin{array}{l}
x = \arccos \left( {\dfrac{1}{4}} \right) + k2\pi \\
x =  - \arccos \left( {\dfrac{1}{4}} \right) + k2\pi 
\end{array} \right.
\end{array}$