Đáp án:
\(\begin{array}{l}
Min = - 1 \Leftrightarrow \sin x = - 1 \Leftrightarrow x = - \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)\\
Max = 3 \Leftrightarrow \sin x = 1 \Leftrightarrow x = \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
y = 2{\sin ^2}x + 2\sin x - 1\\
= {\left( {\sqrt 2 \sin x} \right)^2} + 2\sqrt 2 \sin x.\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{2} - \dfrac{3}{2}\\
= {\left( {\sqrt 2 \sin x + \dfrac{1}{{\sqrt 2 }}} \right)^2} - \dfrac{3}{2}\\
Do: - 1 \le \sin x \le 1\\
\to - \sqrt 2 \le \sqrt 2 \sin x \le \sqrt 2 \\
\to - \sqrt 2 + \dfrac{1}{{\sqrt 2 }} \le \sqrt 2 \sin x + \dfrac{1}{{\sqrt 2 }} \le \sqrt 2 + \dfrac{1}{{\sqrt 2 }}\\
\to \dfrac{1}{2} \le {\left( {\sqrt 2 \sin x + \dfrac{1}{{\sqrt 2 }}} \right)^2} \le \dfrac{9}{2}\\
\to - 1 \le {\left( {\sqrt 2 \sin x + \dfrac{1}{{\sqrt 2 }}} \right)^2} - \dfrac{3}{2} \le 3\\
\to Min = - 1 \Leftrightarrow \sin x = - 1 \Leftrightarrow x = - \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)\\
Max = 3 \Leftrightarrow \sin x = 1 \Leftrightarrow x = \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)
\end{array}\)
