a) $y = 2\sin^2x - \cos2x$
$= 1 - 2\cos2x$
Ta có: $- 1 \leq \cos2x \leq 1$
$\Leftrightarrow -2\leq -2\cos2x \leq 2$
$\Leftrightarrow - 1 \leq 1 - 2\cos2x \leq 3$
Hay $-1 \leq y \leq 3$
Vậy $\min y = - 1 \Leftrightarrow \cos2x = 1\Leftrightarrow x = k\pi$
$\max y = 3 \Leftrightarrow \cos2x = - 1 \Leftrightarrow x = \dfrac{\pi}{2} + k\pi\quad (k \in \Bbb Z)$
b) $y = 3 - 2|\sin x|$
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 $\min y = 1 \Leftrightarrow \sin x = 0 \Leftrightarrow x = k\pi$
$\max y = 3 \Leftrightarrow \sin x = \pm 1 \Leftrightarrow x = \dfrac{\pi}{2} + k\pi \quad (k \in \Bbb Z)$
c) $y = \cos x + \cos\left(x - \dfrac{\pi}{3}\right)$
$= \cos x + \cos x.\cos\dfrac{\pi}{3} + \sin x.\sin\dfrac{\pi}{3}$
$= \dfrac{3}{2}\cos x + \dfrac{\sqrt3}{2}\sin x$
$= \sqrt3.\left(\dfrac{\sqrt3}{2}\cos x + \dfrac{1}{2}\sin x\right)$
$= \sqrt3.\cos\left(x - \dfrac{\pi}{6}\right)$
Ta có:
$-1 \leq \cos\left(x - \dfrac{\pi}{6}\right) \leq 1$
$\Leftrightarrow -\sqrt3 \leq \sqrt3.\cos\left(x - \dfrac{\pi}{6}\right) \leq \sqrt3$
Hay $-\sqrt3 \leq y \leq \sqrt3$
Vậy $\min y = - \sqrt3 \Leftrightarrow \cos\left(x - \dfrac{\pi}{6}\right) = -1 \Leftrightarrow x = \dfrac{7\pi}{6} + k2\pi$
$\max y = \sqrt3 \Leftrightarrow \cos\left(x - \dfrac{\pi}{6}\right) = 1 \Leftrightarrow x = \dfrac{\pi}{6} + k2\pi \quad (k \in \Bbb Z)$
