Đáp án:
$\begin{array}{l}
a)y = 3\sin 2x + 4\cos 2x - 1\\
\Leftrightarrow 3\sin 2x + 4\cos 2x = y + 1\\
Dk:{3^2} + {4^2} \ge {\left( {y + 1} \right)^2}\\
\Leftrightarrow {\left( {y + 1} \right)^2} \le 25\\
\Leftrightarrow - 5 \le y + 1 \le 5\\
\Leftrightarrow - 6 \le y \le 4\\
\Leftrightarrow \left\{ \begin{array}{l}
GTNN:y = - 6\\
GTLN:y = 4
\end{array} \right.\\
b)\\
y = 2\sin x - 3\cos x - 2\\
\Leftrightarrow 2\sin x - 3\cos x = y + 2\\
Dk:{2^2} + {\left( { - 3} \right)^2} \ge {\left( {y + 2} \right)^2}\\
\Leftrightarrow {\left( {y + 2} \right)^2} \le 13\\
\Leftrightarrow - \sqrt {13} \le y + 2 \le \sqrt {13} \\
\Leftrightarrow - \sqrt {13} - 2 \le y \le \sqrt {13} - 2\\
\Leftrightarrow \left\{ \begin{array}{l}
GTNN:y = - \sqrt {13} - 2\\
GTLN:y = \sqrt {13} - 2
\end{array} \right.
\end{array}$
