Đáp án:
$MinY = \dfrac{1}{2} \Leftrightarrow {\sin ^2}2x = 1 \Leftrightarrow \cos 2x = 0 \Leftrightarrow 2x = \dfrac{\pi }{2} + k\pi \Leftrightarrow x = \dfrac{\pi }{4} + k\dfrac{\pi }{2}\left( {k \in Z} \right)$
$MaxY = 1 \Leftrightarrow {\sin ^2}2x = 0 \Leftrightarrow \sin 2x = 0 \Leftrightarrow 2x = k\pi \Leftrightarrow x = k\dfrac{\pi }{2}\left( {k \in Z} \right)$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
Y = {\sin ^4}x + {\cos ^4}x\\
= {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x\\
= {1^2} - \dfrac{{{{\sin }^2}2x}}{2}\\
= 1 - \dfrac{{{{\sin }^2}2x}}{2}
\end{array}$
Mà $0 \le {\sin ^2}2x \le 1 \Rightarrow \dfrac{{ - 1}}{2} \le \dfrac{{ - {{\sin }^2}2x}}{2} \le 0$
$ \Rightarrow \dfrac{1}{2} \le Y \le 1$
Như vậy:;
$MinY = \dfrac{1}{2} \Leftrightarrow {\sin ^2}2x = 1 \Leftrightarrow \cos 2x = 0 \Leftrightarrow 2x = \dfrac{\pi }{2} + k\pi \Leftrightarrow x = \dfrac{\pi }{4} + k\dfrac{\pi }{2}\left( {k \in Z} \right)$
$MaxY = 1 \Leftrightarrow {\sin ^2}2x = 0 \Leftrightarrow \sin 2x = 0 \Leftrightarrow 2x = k\pi \Leftrightarrow x = k\dfrac{\pi }{2}\left( {k \in Z} \right)$
