y=2sin^2+2sinx-1 tìm GTLN và GTNN

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:

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