Đáp án:
$\begin{cases}\min y = \dfrac{1}{2} \Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\\\max y = 1 \Leftrightarrow x = k\dfrac{\pi}{2}\end{cases} \quad (k \in \Bbb Z)$
Giải thích các bước giải:
$y = \sin^4x + \cos^4x$
$\to y = (\sin^2x + \cos^2x)^2 - 2\sin^2x\cos^2x$
$\to y = 1 - \dfrac{1}{2}\sin^22x$
$\to y = 1 - \dfrac{1}{4}(1 - \cos4x)$
$\to y = \dfrac{3 + \cos4x}{4}$
Ta có:
$-1 \leq \cos4x \leq 1$
$\to 2 \leq 3 + \cos4x \leq 4$
$\to \dfrac{1}{2} \leq \dfrac{3 + \cos4x}{4} \leq 1$
$\to \dfrac{1}{2} \leq y \leq 1$
Vậy $\min y = \dfrac{1}{2} \Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}$
$\max y = 1 \Leftrightarrow x = k\dfrac{\pi}{2} \quad (k \in \Bbb Z)$
