$y = 8 + \cos^2x$
Ta có: $0 \leq \cos^2x \leq 1$
$\Leftrightarrow 8 \leq 8 + \cos^2x \leq 9$
Vậy $\min y = 8 \Leftrightarrow \cos x = 0 \Leftrightarrow x = \dfrac{\pi}{2} + k\pi$
$\max y = 9 \Leftrightarrow \cos^2x = 1 \Leftrightarrow \cos x = \pm 1 \Leftrightarrow x = k\pi \quad (k \in \Bbb Z)$
Cách khác:
$y = 8 + \cos^2x$
$= 8 + \dfrac{1 + \cos2x}{2}$
$= \dfrac{17 + \cos2x}{2}$
Ta có: $-1 \leq \cos2x \leq 1$
$\Leftrightarrow 16 \leq 17 + \cos2x\leq 18$
$\Leftrightarrow 8 \leq \dfrac{17 + \cos2x}{2} \leq 9$
Vậy $\min y = 8; \,\max y = 9$
