Đáp án:
$\begin{cases}miny = 5 \Leftrightarrow x=\dfrac{\pi}4+\dfrac{k\pi}2\\maxy = 6 \Leftrightarrow x=\dfrac{k\pi}2\end{cases}\,\,\,(k \in \Bbb Z)$
Giải thích các bước giải:
$6 - 4sin^2x.cos^2x$
$= 6 - (2sinx.cosx)^2$
$=6 - sin^22x$
$= 6 - \dfrac{1 - cos4x}{2}$
$= \dfrac{11 + cos4x}{2}$
Ta có:
$-1 \leq cos4x \leq 1$
$\Leftrightarrow 10 \leq 11 + cos4x \leq 12$
$\Leftrightarrow 5 \leq \dfrac{11 + cos4x}{2} \leq 6$
Hay $5 \leq y \leq 6$
Vậy $miny = 5$ khi $\cos 4x=-1\Leftrightarrow x=\dfrac{\pi}4+\dfrac{k\pi}2$
$\, maxy = 6$ khi $\cos 4x=1\Leftrightarrow x=\dfrac{k\pi}2\,\,\, (k \in \Bbb Z)$
