\[\begin{array}{l}
y = 2{\sin ^2}\left( {\frac{{2x + 3\pi }}{4}} \right)\,\,,\,\,\,x \in \left[ {\pi ;\,\,\frac{{9\pi }}{4}} \right]\\
Ta\,\,co:\,\,\,0 \le {\sin ^2}\left( {\frac{{2x + 3\pi }}{4}} \right) \le 1\\
\Rightarrow 0 \le 2{\sin ^2}\left( {\frac{{2x + 3\pi }}{4}} \right) \le 2\\
\Rightarrow \mathop {Min}\limits_{\left[ {\pi ;\,\,\frac{{9\pi }}{4}} \right]} \,y = 0\,\,\,khi\,\,\,\sin \left( {\frac{{2x + 3\pi }}{4}} \right) = 0 \Leftrightarrow \frac{{2x + 3\pi }}{4} = k\pi \\
\Leftrightarrow 2x + 3\pi = 4k\pi \\
\Leftrightarrow x = \frac{{ - 3\pi }}{2} + 2k\pi \\
Ma\,\,\,x \in \left[ {\pi ;\,\,\frac{{9\pi }}{4}} \right] \Rightarrow \pi \le \frac{{ - 3\pi }}{2} + 2k\pi \le \frac{{9\pi }}{4}\\
\Rightarrow x \in \left\{ {.....} \right\}\\
\mathop {Max}\limits_{\left[ {\pi ;\,\,\frac{{9\pi }}{4}} \right]} \,\,\,\,y = 2\,\,\,khi\,\,\sin \left( {\frac{{2x + 3\pi }}{4}} \right) = 1 \Leftrightarrow \cos \left( {\frac{{2x + 3\pi }}{4}} \right) = 0
\end{array}\]
Đến đây em tìm x và làm tương tự với TH Max nhé em.
