Đáp án:
\[\left\{ \begin{array}{l}
{y_{\min }} = - 3 \Leftrightarrow x = \pi + k2\pi \\
{y_{\max }} = 2\sqrt 2 - 3 \Leftrightarrow x = k2\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
\(\begin{array}{l}
y = 2\sqrt {\cos x + 1} - 3\\
- 1 \le \cos x \le 1,\,\,\,\forall x\\
\Leftrightarrow 0 \le \cos x + 1 \le 2,\,\,\,\forall x\\
\Rightarrow 0 \le \sqrt {\cos x + 1} \le \sqrt 2 ,\,\,\,\forall x\\
\Leftrightarrow 0 \le 2\sqrt {\cos x + 1} \le 2\sqrt 2 ,\,\,\,\forall x\\
\Leftrightarrow - 3 \le 2\sqrt {\cos x + 1} - 3 \le 2\sqrt 2 - 3,\,\,\,\forall x\\
\Leftrightarrow - 3 \le y \le 2\sqrt 2 - 3\\
\Rightarrow \left\{ \begin{array}{l}
{y_{\min }} = - 3 \Leftrightarrow \cos x = - 1 \Leftrightarrow x = \pi + k2\pi \\
{y_{\max }} = 2\sqrt 2 - 3 \Leftrightarrow \cos x = 1 \Leftrightarrow x = k2\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
Vậy \(\left\{ \begin{array}{l}
{y_{\min }} = - 3 \Leftrightarrow x = \pi + k2\pi \\
{y_{\max }} = 2\sqrt 2 - 3 \Leftrightarrow x = k2\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\)
