Tham khảo nhé :

a.

$y=4sin^2x+8sinx-1$

Đặt $sinx=t$ với $-1 \leq t \leq 1$

Khảo sát hàm $y=4t^2+8t-1$ trên $[-1;1]$ được:

$ \begin{array}{c|ccc} t & -1 & & 1 \\\hline & & & 11 \\ & & \nearrow & \\ y & -5 & & \end{array}$

Vậy $-5 \leq y \leq 11$

b.

`y=\sqrt{3-sin^2 2x}\Rightarrow \sqrt{2}\leq y\leq \sqrt{3}(0 \leq sin^2 2x \leq 1)`

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 $\min y = \sqrt2 \Leftrightarrow \sin2x = \pm 1 \Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}$

$\max y = \sqrt3 \Leftrightarrow \sin2x = 0 \Leftrightarrow x = k\dfrac{\pi}{2}\quad (k\in\Bbb Z)$