Đáp án:
\(Min = 3\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:\left\{ \begin{array}{l}
\cos \left( {x + \dfrac{\pi }{4}} \right) \ne 0\\
\sin x \ne - \dfrac{1}{2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x + \dfrac{\pi }{4} \ne \dfrac{\pi }{2} + k\pi \\
x \ne - \dfrac{\pi }{6} + k2\pi \\
x \ne \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right. \to \left\{ \begin{array}{l}
x \ne \dfrac{\pi }{4} + k\pi \\
x \ne - \dfrac{\pi }{6} + k2\pi \\
x \ne \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
b)Do:0 \le {\cos ^2}x \le 1\\
\to 0 \ge - 2{\cos ^2}x \ge - 2\\
\to 5 \ge 5 - 2{\cos ^2}x \ge 3\\
\to Min = 3 \Leftrightarrow {\cos ^2}x = 1 \to {\sin ^2}x = 0\\
\to \sin x = 0 \to x = k\pi \left( {k \in Z} \right)\\
Max = 5 \Leftrightarrow {\cos ^2}x = 0 \to \cos x = 0 \to x = \dfrac{\pi }{2} + k\pi \left( {k \in Z} \right)
\end{array}\)
