Đáp án:
$\begin{array}{l}
1)y = 3 - \sin x.\cos x\\
= 3 - \dfrac{1}{2}.2.\sin x.\cos x\\
= 3 - \dfrac{1}{2}.\sin 2x\\
Do: - 1 \le \sin 2x \le 1\\
\Rightarrow - \dfrac{1}{2} \le - \dfrac{1}{2}\sin 2x \le \dfrac{1}{2}\\
\Rightarrow \dfrac{5}{2} \le y \le \dfrac{7}{2}\\
\Rightarrow \left\{ \begin{array}{l}
GTLN:y = \dfrac{7}{2}\\
khi:\sin 2x = - 1 \Rightarrow x = \dfrac{{ - \pi }}{4} + k\pi \\
GTNN:y = \dfrac{5}{2}\\
khi:\sin 2x = 1 \Rightarrow x = \dfrac{\pi }{2} + k\pi
\end{array} \right.\\
2)y = {\sin ^2}x - {\cos ^2}x\\
= - \left( {{{\cos }^2}x - {{\sin }^2}x} \right)\\
= - \cos 2x\\
\Rightarrow - 1 \le y \le 1\\
\Rightarrow \left\{ \begin{array}{l}
GTNN:y = - 1\,\\
Khi:\cos 2x = 1 \Rightarrow x = \dfrac{\pi }{2} + k\pi \\
GTLN:y = 1\\
Khi:x = k\pi
\end{array} \right.\\
3)y = 1 - 3{\cos ^2}x\\
Do:0 \le {\cos ^2}x \le 1\\
\Rightarrow - 3 \le - 3{\cos ^2}x \le 0\\
\Rightarrow - 2 \le y \le 1\\
\Rightarrow \left\{ \begin{array}{l}
GTNN:y = - 2\\
\,khi:{\cos ^2}x = 1 \Rightarrow x = k\pi \\
GTLN:y = 1\\
Khi:{\cos ^2}x = 0 \Rightarrow x = \dfrac{\pi }{2} + k\pi
\end{array} \right.\\
4)y = \dfrac{2}{{3 + \cos x}}\\
\Rightarrow 2 \le 3 + \cos x \le 4\\
\Rightarrow \dfrac{2}{4} \le \dfrac{2}{{3 + \cos x}} \le \dfrac{2}{2}\\
\Rightarrow \dfrac{1}{2} \le y \le 1\\
\Rightarrow \left\{ \begin{array}{l}
gtnn:y = \dfrac{1}{2}\\
GTLN:y = 1
\end{array} \right.\\
5)y = \sin x + \cos x\\
= \sqrt 2 .\sin \left( {x + \dfrac{\pi }{4}} \right)\\
\Rightarrow - \sqrt 2 \le y \le \sqrt 2 \\
6)y = 2{\sin ^2}x - \cos 2x\\
= - \left( {1 - 2{{\sin }^2}x} \right) - \cos 2x + 1\\
= - \cos 2x - \cos 2x + 1\\
= - 2\cos 2x + 1\\
\Rightarrow - 1 \le y \le 3\\
\Rightarrow \left\{ \begin{array}{l}
GTNN:y = - 1\\
GTLN:y = 3
\end{array} \right.\\
7)y = 2\left| {\cos x} \right| - 3\\
Do:0 \le \left| {\cos x} \right| \le 1\\
\Rightarrow - 3 \le y \le - 2\\
8)y = \cos x - \sqrt 3 \sin x\\
y = 2.\left( {\dfrac{1}{2}.\cos x - \dfrac{{\sqrt 3 }}{2}\sin x} \right)\\
\Rightarrow y = 2.\cos \left( {x + \dfrac{\pi }{3}} \right)\\
\Rightarrow - 2 \le y \le 2
\end{array}$