a) $y = -4\cos^2x + 8\sin x + 3$
$\to y = -4(1 -\sin^2x) + 8\sin x + 3$
$\to y = 4\sin^2x + 8\sin x - 1$
$\to y = 4(\sin x + 1)^2- 5$
Ta có:
$-1\leq \sin x \leq 1$
$\to 0 \leq \sin x + 1 \leq 2$
$\to 0 \leq (\sin x + 1)^2 \leq 4$
$\to 0 \leq 4(\sin x + 1)^2 \leq 16$
$\to - 5\leq 4(\sin x + 1)^2 - 5 \leq 11$
Hay $-5\leq y \leq 11$
Vậy $\min y = - 5 \Leftrightarrow \sin x = -1\Leftrightarrow x = -\dfrac{\pi}{2} + k2\pi$
$\max y = 11 \Leftrightarrow \sin x = 1 \Leftrightarrow x = \dfrac{\pi}{2} + k2\pi\quad (k\in\Bbb Z)$
b) $y =\sqrt{3 - 4\sin^2x\cos^2x}$
$\to y = \sqrt{3 - \sin^22x}$
Ta có:
$0 \leq \sin^22x \leq 1$
$\to - 1 \leq -\sin^22x \leq 0$
$\to 2 \leq 3 - \sin^22x \leq 3$
$\to \sqrt2 \leq \sqrt{3 - \sin^22x} \leq \sqrt3$
Hay $\sqrt2 \leq y \leq \sqrt3$
Vậy $\min y = \sqrt2 \Leftrightarrow \sin2x = \pm 1 \Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}$
$\max y = \sqrt3 \Leftrightarrow \sin2x = 0 \Leftrightarrow x = k\dfrac{\pi}{2}\quad (k\in\Bbb Z)$
