Đáp án:
$\begin{array}{l}
c)y = \frac{1}{{\sqrt {{\mathop{\rm s}\nolimits} {\rm{inx}} + 1} }}\\
Dkxd:\sin x > - 1\\
Do: - 1 < \sin x \le 1\\
\Rightarrow 0 < \sqrt {{\mathop{\rm sinx}\nolimits} + 1} \le \sqrt 2 \\
\Rightarrow \frac{1}{{\sqrt {\sin x + 1} }} \ge \frac{1}{{\sqrt 2 }}\\
\Rightarrow y \ge \frac{1}{{\sqrt 2 }}\\
\Rightarrow GTNN:y = \frac{1}{{\sqrt 2 }} \Leftrightarrow \sin x = 1\\
\Rightarrow x = \frac{\pi }{2} + k2\pi \\
d)0 \le co{s^2}x \le 1\\
\Rightarrow - 1 \le - {\cos ^2}x \le 0\\
\Rightarrow 0 \le 1 - {\cos ^2}x \le 1\\
\Rightarrow 0 \le \sqrt {1 - {{\cos }^2}x} \le 1\\
\Rightarrow 0 \le 1 - \sqrt {1 - {{\cos }^2}x} \le 1\\
\Rightarrow \left\{ \begin{array}{l}
GTNN:y = 0 \Leftrightarrow {\cos ^2}x = 0\\
GTLN:y = 1 \Leftrightarrow {\cos ^2}x = 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
GTNN:y = 0 \Leftrightarrow x = \frac{\pi }{2} + k\pi \\
GTLN:y = 1 \Leftrightarrow x = k\pi
\end{array} \right.
\end{array}$
