Đáp án:
$\begin{array}{l}
y = \dfrac{{\sin x - 1}}{{2\sin x + \cos x + 3}}\\
\Rightarrow 2y\sin x + y.\cos x + 3y = \sin x - 1\\
\Rightarrow \left( {2y - 1} \right)\sin x + y.\cos x = - 3y - 1\\
Dk:{\left( {2y - 1} \right)^2} + {y^2} \ge {\left( { - 3y - 1} \right)^2}\\
\Rightarrow 5{y^2} - 4y + 1 \ge 9{y^2} + 6y + 1\\
\Rightarrow 4{y^2} + 10y \le 0\\
\Rightarrow 2y\left( {2y + 5} \right) \le 0\\
\Rightarrow - \dfrac{5}{2} \le y \le 0\\
\Rightarrow \left\{ \begin{array}{l}
GTNN:y = - \dfrac{5}{2}\\
GTLN:y = 0
\end{array} \right.
\end{array}$
