Tìm GTLN và GTNN của các hàm số sau:

1) $y = \dfrac{\sin x\cos x +3}{4}$

$= \dfrac{\sin2x +6}{8}$

Ta có: $-1 \leq \sin2x \leq 1$

$\Leftrightarrow 5 \leq \sin2x + 6 \leq 7$

$\Leftrightarrow \dfrac{5}{8} \leq \dfrac{\sin2x +6}{8} \leq \dfrac{7}{8}$

Hay $\dfrac{5}{8} \leq y \dfrac{7}{8}$

Vậy $\min y = \dfrac{5}{8} \Leftrightarrow \sin2x = -1 \Leftrightarrow x = -\dfrac{\pi}{4} + k\pi$

$\max y = \dfrac{7}{8} \Leftrightarrow \sin2x = 1 \Leftrightarrow x = \dfrac{\pi}{4} + k\pi \quad (k \in \Bbb Z)$

2) $y = \dfrac{3\cos^2x -1}{2}$

$= \dfrac{3(1 + \cos2x) - 2}{4}$

$= \dfrac{3\cos2x + 1}{4}$

Ta có: $-1 \leq \cos2x \leq 1$

$\Leftrightarrow -3 \leq 3\cos2x \leq 3$

$\Leftrightarrow -2 \leq 3\cos2x + 1 \leq 4$

$\Leftrightarrow -\dfrac{1}{2} \leq \dfrac{3\cos2x +1}{4} \leq 1$

Hay $-\dfrac{1}{2} \leq y \leq 1$

Vậy $\min y = -\dfrac{1}{2} \Leftrightarrow \cos2x = -1 \Leftrightarrow x = \dfrac{\pi}{2} + k\pi$

$\max y = 1 \Leftrightarrow \cos2x = 1 \Leftrightarrow x = k\pi \quad (k \in \Bbb Z)$