Đáp án:
$\begin{array}{l}
a)y = 3 - 2\cos x\\
Do: - 1 \le \cos x \le 1\\
\Leftrightarrow - 2 \le - 2\cos x \le 2\\
\Leftrightarrow 1 \le 3 - 2\cos x \le 5\\
\Leftrightarrow 1 \le y \le 5\\
\Leftrightarrow \left\{ \begin{array}{l}
GTNN:y = 1\,khi:\cos x = 1 \Leftrightarrow x = k2\pi \\
GTLN:y = 5\,khi:\cos x = - 1 \Leftrightarrow x = \pi + k2\pi
\end{array} \right.\\
b)\\
y = 3\sin x - 1\\
Do: - 1 \le \sin x \le 1\\
\Leftrightarrow - 3 \le 3\sin x \le 3\\
\Leftrightarrow - 4 \le 3\sin x - 1 \le 2\\
\Leftrightarrow - 4 \le y \le 2\\
\Leftrightarrow \left\{ \begin{array}{l}
GTNN:y = - 4\,khi:\sin x = - 1 \Leftrightarrow x = - \dfrac{\pi }{2} + k2\pi \\
GTLN:y = 2\,khi:\sin x = 1 \Leftrightarrow x = \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
c)\\
y = {\sin ^2}x + 5\\
Do: - 1 \le \sin x \le 1\\
\Leftrightarrow 0 \le {\sin ^2}x \le 1\\
\Leftrightarrow 5 \le {\sin ^2}x + 5 \le 6\\
\Leftrightarrow 5 \le y \le 6\\
\Leftrightarrow \left\{ \begin{array}{l}
GTNN:y = 5\,khi:\sin x = 0 \Leftrightarrow x = k\pi \\
GTLN:y = 6\,khi:\sin x = \pm 1 \Leftrightarrow x = \dfrac{\pi }{2} + k\pi
\end{array} \right.
\end{array}$
