Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
- 1 \le \cos x \le 1 \Rightarrow - 4 \le 4\cos x \le 4 \Rightarrow - 2 \le 2 + 4\cos x \le 6\\
\Rightarrow {y_{\min }} = - 2 \Leftrightarrow \cos x = - 1 \Leftrightarrow x = \pi + k2\pi \\
{y_{\max }} = 6 \Leftrightarrow \cos x = 1 \Leftrightarrow x = k2\pi \\
4,\\
y = 2{\sin ^2}x - \cos 2x = 2{\sin ^2}x - \left( {1 - 2{{\sin }^2}x} \right) = 4{\sin ^2}x - 1\\
- 1 \le \sin x \le 1 \Rightarrow 0 \le {\sin ^2}x \le 1 \Rightarrow - 1 \le 4{\sin ^2}x - 1 \le 3\\
\Rightarrow {y_{\min }} = - 1 \Leftrightarrow \sin x = 0 \Leftrightarrow x = k\pi \\
{y_{\max }} = 3 \Leftrightarrow \sin x = \pm 1 \Leftrightarrow x = \dfrac{\pi }{2} + k\pi \\
6,\\
y = \cos x + \cos \left( {x - \dfrac{\pi }{3}} \right)\\
= \cos x + \cos x.\cos \dfrac{\pi }{3} + \sin x.\sin \dfrac{\pi }{3}\\
= \dfrac{3}{2}\cos x + \dfrac{{\sqrt 3 }}{2}\sin x\\
y = a.\sin x + b.\cos x \Rightarrow - \sqrt {{a^2} + {b^2}} \le y \le \sqrt {{a^2} + {b^2}} \\
\Rightarrow - \sqrt {{{\left( {\dfrac{3}{2}} \right)}^2} + {{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2}} \le y \le \sqrt {{{\left( {\dfrac{3}{2}} \right)}^2} + {{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2}} \\
\Leftrightarrow - \sqrt 3 \le y \le \sqrt 3 \\
7,\\
y = {\cos ^2}x + 2\cos 2x = {\cos ^2}x + 2.\left( {2{{\cos }^2}x - 1} \right) = 5{\cos ^2}x - 2\\
- 1 \le \cos x \le 1 \Rightarrow 0 \le {\cos ^2}x \le 1 \Leftrightarrow - 2 \le 5{\cos ^2}x - 2 \le 3\\
\Rightarrow {y_{\min }} = - 2 \Leftrightarrow \cos x = 0 \Leftrightarrow x = \dfrac{\pi }{2} + k\pi \\
{y_{\max }} = 3 \Leftrightarrow \cos x = \pm 1 \Leftrightarrow x = k\pi
\end{array}\)
