$y = 1 - 2\sin3x.\cos3x$
$= 1 - \sin6x$
Ta có:
$-1 \leq \sin6x \leq 1$
$\Leftrightarrow 0 \leq 1 - \sin6x \leq 2$
Hay $0 \leq y \leq 2$
Vậy $\min y = 0 \Leftrightarrow \sin6x = 1 \Leftrightarrow x = \dfrac{\pi}{12} + k\dfrac{\pi}{3}$
$\max y = 2 \Leftrightarrow \sin6x = -1 \Leftrightarrow x = -\dfrac{\pi}{12} + k\dfrac{\pi}{3} \quad (k \in \Bbb Z)$
