Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
 - 1 \le \cos x \le 1 \Rightarrow  - 2 \le 2\cos x \le 2\\
 \Rightarrow  - 3 \le 2\cos x - 1 \le 1\\
 \Rightarrow \left\{ \begin{array}{l}
{y_{\min }} =  - 3 \Leftrightarrow \cos x =  - 1 \Leftrightarrow x = \pi  + k2\pi \\
{y_{\max }} = 1 \Leftrightarrow \cos x = 1 \Leftrightarrow x = k2\pi 
\end{array} \right.\\
d,\\
y = 2{\sin ^2}x + 1\\
 - 1 \le \sin \,x \le 1 \Rightarrow 0 \le {\sin ^2}x \le 1\\
 \Leftrightarrow 1 \le 2{\sin ^2}x + 1 \le 3\\
 \Rightarrow \left\{ \begin{array}{l}
{y_{\min }} = 1 \Leftrightarrow {\sin ^2}x = 0 \Leftrightarrow \sin x = 0 \Leftrightarrow x = k\pi \\
{y_{\max }} = 3 \Leftrightarrow {\sin ^2}x = 1 \Leftrightarrow \sin x =  \pm 1 \Leftrightarrow x = \dfrac{\pi }{2} + k\pi 
\end{array} \right.\\
g,\\
 - 1 \le \sin \left( {x - \dfrac{\pi }{2}} \right) \le 1\\
 \Rightarrow  - 2 \le 2\sin \left( {x - \dfrac{\pi }{2}} \right) \le 2\\
 \Rightarrow 1 \le 2\sin \left( {x - \dfrac{\pi }{2}} \right) + 3 \le 5\\
 \Rightarrow \left\{ \begin{array}{l}
{y_{\min }} = 1 \Leftrightarrow \sin \left( {x - \dfrac{\pi }{2}} \right) =  - 1 \Leftrightarrow x - \dfrac{\pi }{2} =  - \dfrac{\pi }{2} + k2\pi  \Leftrightarrow x = k2\pi \\
{y_{\max }} = 5 \Leftrightarrow \sin \left( {x - \dfrac{\pi }{2}} \right) = 1 \Leftrightarrow x - \dfrac{\pi }{2} = \dfrac{\pi }{2} + k2\pi  \Leftrightarrow x = \pi  + k2\pi 
\end{array} \right.\\
j,\\
 - 1 \le \sin x \le 1 \Rightarrow 0 \le \left| {\sin x} \right| \le 1\\
 \Rightarrow 0 \le 2\left| {\sin x} \right| \le 2\\
 \Leftrightarrow 1 \le 3 - 2\left| {\sin x} \right| \le 3\\
 \Rightarrow \left\{ \begin{array}{l}
{y_{\min }} = 1 \Leftrightarrow \left| {\sin x} \right| = 1 \Leftrightarrow \sin x =  \pm 1 \Leftrightarrow x = \dfrac{\pi }{2} + k\pi \\
{y_{\max }} = 3 \Leftrightarrow \left| {\sin x} \right| = 0 \Leftrightarrow \sin x = 0 \Leftrightarrow x = k\pi 
\end{array} \right.\\
m,\\
 - 1 \le cos\left( {2x - \dfrac{\pi }{3}} \right) \le 1\\
 \Rightarrow 0 \le {\cos ^2}\left( {2x - \dfrac{\pi }{3}} \right) \le 1\\
 \Rightarrow  - 5 \le  - 5 + 2{\cos ^2}\left( {2x - \dfrac{\pi }{3}} \right) \le  - 3\\
 \Rightarrow \left\{ \begin{array}{l}
{y_{\min }} =  - 5 \Leftrightarrow \cos \left( {2x - \dfrac{\pi }{3}} \right) = 0 \Leftrightarrow 2x - \dfrac{\pi }{3} = \dfrac{\pi }{2} + k\pi  \Leftrightarrow x = \dfrac{{5\pi }}{{12}} + \dfrac{{k\pi }}{2}\\
{y_{\max }} =  - 3 \Leftrightarrow \cos \left( {2x - \dfrac{\pi }{3}} \right) =  \pm 1 \Leftrightarrow 2x - \dfrac{\pi }{3} = k\pi  \Leftrightarrow x = \dfrac{\pi }{6} + \dfrac{{k\pi }}{2}
\end{array} \right.
\end{array}\)