Đáp án:
$\begin{cases}\min y = 1 \Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\\\max y = 3\Leftrightarrow x = k\dfrac{\pi}{2}\end{cases}\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\quad y = 3 - 8\sin^2x\cos^2x$
$\Leftrightarrow y = 3 - 2\sin^22x$
$\Leftrightarrow y = 3 - (1-\cos4x)$
$\Leftrightarrow y = 2 + \cos4x$
Ta có:
$\quad -1\leqslant \cos4x \leqslant 1$
$\Leftrightarrow 1 \leqslant 2 +\cos4x \leqslant 3$
$\Leftrightarrow 1 \leqslant y \leqslant 3$
Do đó:
$+)\quad \min y = 1$
$\Leftrightarrow \cos4x = -1$
$\Leftrightarrow 4x = \pi + k2\pi$
$\Leftrightarrow x =\dfrac{\pi}{4} + k\dfrac{\pi}{2}\quad (k\in\Bbb Z)$
$+)\quad \max y = 3$
$\Leftrightarrow \cos4x = 1$
$\Leftrightarrow 4x = k2\pi$
$\Leftrightarrow x =k\dfrac{\pi}{2}\quad (k\in\Bbb Z)$
Vậy $\begin{cases}\min y = 1 \Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\\\max y = 3\Leftrightarrow x = k\dfrac{\pi}{2}\end{cases}\quad (k\in\Bbb Z)$
