Giải thích các bước giải:

\(\begin{array}{l}
a,\\
y = 2{\sin ^2}x - \cos 2x = 2{\sin ^2}x - \left( {1 - 2{{\sin }^2}x} \right) = 4{\sin ^2}x - 1\\
 - 1 \le \sin x \le 1 \Rightarrow 0 \le {\sin ^2}x \le 1 \Rightarrow  - 1 \le 4{\sin ^2}x - 1 \le 3\\
 \Rightarrow {y_{\min }} =  - 1 \Leftrightarrow \sin x = 0 \Leftrightarrow x = k\pi \\
{y_{\max }} = 3 \Leftrightarrow \sin x =  \pm 1 \Leftrightarrow x = \dfrac{\pi }{2} + k\pi \backslash \\
b,\\
y = {\sin ^4}x + {\cos ^4}x + 4 = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x.{\cos ^2}x + 4\\
 = {1^2} - \dfrac{1}{2}.{\left( {2\sin x.\cos x} \right)^2} + 4\\
 = 5 - \dfrac{1}{2}{\sin ^2}2x\\
 - 1 \le \sin 2x \le 1 \Rightarrow 0 \le {\sin ^2}2x \le 1\\
 \Rightarrow 4 \le 5 - \dfrac{1}{2}{\sin ^2}2x \le \dfrac{9}{2}\\
 \Rightarrow {y_{\min }} = 4 \Leftrightarrow \sin 2x =  \pm 1 \Leftrightarrow 2x = \dfrac{\pi }{2} + k\pi  \Leftrightarrow x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
{y_{\max }} = \dfrac{9}{2} \Leftrightarrow \sin 2x = 0 \Leftrightarrow 2x = k\pi  \Leftrightarrow x = \dfrac{{k\pi }}{2}\\
c,\\
y = \cos x + \cos \left( {x - \dfrac{\pi }{3}} \right)\\
 = \cos x + \cos x.\cos \dfrac{\pi }{3} + \sin x.\sin \dfrac{\pi }{3}\\
 = \dfrac{3}{2}\cos x + \dfrac{{\sqrt 3 }}{2}\sin x\\
y = a.\sin x + b.\cos x \Rightarrow  - \sqrt {{a^2} + {b^2}}  \le y \le \sqrt {{a^2} + {b^2}} \\
 \Rightarrow  - \sqrt {{{\left( {\dfrac{3}{2}} \right)}^2} + {{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2}}  \le y \le \sqrt {{{\left( {\dfrac{3}{2}} \right)}^2} + {{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2}} \\
 \Leftrightarrow  - \sqrt 3  \le y \le \sqrt 3 
\end{array}\)