a) $y = \cos^2x + 2\cos x$

$= (\cos x + 1)^2 - 1$

Ta có: $-1 \leq \cos x \leq 1$

$\Leftrightarrow 0 \leq \cos x + 1 \leq 2$

$\Leftrightarrow 0 \leq (\cos x + 1)^2 \leq 4$

$\Leftrightarrow - 1 \leq (\cos x + 1)^2 - 1 \leq 3$

Hay $-1 \leq y \leq 3$

Vậy $\min y = - 1 \Leftrightarrow \cos x = -1 \Leftrightarrow x = \pi + k2\pi$

$\max y = 3 \Leftrightarrow \cos x = 1 \Leftrightarrow k2\pi \quad (k\ in \Bbb Z)$

b) $y = \sqrt{5 - 2\cos^2x\sin^2x}$

$= \sqrt{5 - \dfrac{1}{2}\sin^22x}$

Ta có: $0 \leq \sin^22x \leq 1$

$\Leftrightarrow 0 \leq \dfrac{1}{2}\sin^22x \leq \dfrac{1}{2}$

$\Leftrightarrow - \dfrac{1}{2} \leq -\dfrac{1}{2}\sin^22x \leq 0$

$\Leftrightarrow \dfrac{9}{2} \leq 5 - \dfrac{1}{2}\sin^22x \leq 5$

$\Leftrightarrow \dfrac{3\sqrt2}{2} \leq \sqrt{5 - \dfrac{1}{2}\sin^22x} \leq \sqrt5$

Hay $\dfrac{3\sqrt2}{2} \leq y \leq \sqrt5$

Vậy $\min y = \dfrac{3\sqrt2}{2} \Leftrightarrow \sin^22x = 1 \Leftrightarrow \sin2x = \pm 1 \Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}$

$\max y = \sqrt5 \Leftrightarrow \sin^22x = 0 \Leftrightarrow \sin2x = 0 \Leftrightarrow x = k\dfrac{\pi}{2}\quad (k \in \Bbb Z)$

c) $y = 3 - 2|\sin x|$

Ta có: $0 \leq |\sin x| \leq 1$

$\Leftrightarrow -2 \leq -2|\sin x| \leq 0$

$\Leftrightarrow 1 \leq 3 - 2|\sin x| \leq 3$

Hay $ 1 \leq y \leq 3$

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

$\max y = 3 \Leftrightarrow \sin x = 0 \Leftrightarrow x = k\pi \quad (k \in \Bbb Z)$

d) $y = \cos x - \cos\left(x - \dfrac{\pi}{3}\right)$

$= \cos x - \cos x.\cos\dfrac{\pi}{3} - \sin x.\sin\dfrac{\pi}{3}$

$= \cos x - \dfrac{1}{2}\cos x - \dfrac{\sqrt3}{2}\sin x$

$= \dfrac{1}{2}\cos x - \dfrac{\sqrt3}{2}\sin x$

$= \cos\left(x + \dfrac{\pi}{3}\right)$

Ta có: $- 1\leq \cos\left(x + \dfrac{\pi}{3}\right) \leq 1$

Hay $ - 1 \leq y \leq 1$

Vậy $\min y = - 1 \Leftrightarrow \cos\left(x + \dfrac{\pi}{3}\right) =  -1 \Leftrightarrow x = \dfrac{2\pi}{3} +k2\pi$

$\max y = 1 \Leftrightarrow \cos\left(x + \dfrac{\pi}{3}\right) = 1 \Leftrightarrow x = - \dfrac{\pi}{3} + k2\pi \quad (k \in \Bbb Z)$