Đáp án:
$\begin{array}{l}
a) - 1 \le \sin x \le 1\\
\Rightarrow - 9 \le - 9\sin x \le 9\\
\Rightarrow - 9 \le - 9\sin x + 1 \le 10\\
\Rightarrow - 9 \le y \le 10\\
\Rightarrow \left\{ \begin{array}{l}
GTNN:y = - 9\\
GTLN:y = 10
\end{array} \right.\\
b) - 1 \le \cos x \le 1\\
\Rightarrow - 2 \le - 2\cos x \le 2\\
\Rightarrow - 1 \le 1 - 2\cos x \le 3\\
\Rightarrow - 1 \le y \le 3\\
\Rightarrow \left\{ \begin{array}{l}
GTNN:y = - 1\\
GTLN:y = 3
\end{array} \right.\\
c) - 1 \le \sin 2x \le 1\\
\Rightarrow - 6 \le 4\sin x - 2 \le 2\\
\Rightarrow \left\{ \begin{array}{l}
GTNN:y = - 6\\
GTLN:y = 2
\end{array} \right.\\
D5)\\
a)2{\sin ^2}x - 3\sin x + 1 = 0\\
\Rightarrow \left( {2\sin x - 1} \right)\left( {\sin x - 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sin x = \dfrac{1}{2}\\
\sin x = 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi \\
x = \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
b)2{\cos ^2}x - 3\cos x - 2 = 0\\
\Rightarrow \left( {2\cos x + 1} \right)\left( {\cos x - 2} \right) = 0\\
\Rightarrow \cos x = - \dfrac{1}{2}\left( {do:\cos x < 2} \right)\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{{2\pi }}{3} + k2\pi \\
x = - \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\\
c)2{\tan ^2}x - 2\tan x - 4 = 0\\
\Rightarrow {\tan ^2}x - \tan x - 2 = 0\\
\Rightarrow \left( {\tan x - 2} \right)\left( {\tan x + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\tan x = 2\\
\tan x = - 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \arctan 2 + k\pi \\
x = - \dfrac{\pi }{4} + k\pi
\end{array} \right.
\end{array}$
