$y = (3\cos x - 4\sin x)(4\cos x - 3\sin x)$
$= 12\cos^2x - 9\sin x\cos x - 16\sin x\cos x + 12\sin^2x$
$= 12 - \dfrac{25}{2}\sin2x$
Ta có: $-1 \leq \sin2x \leq 1$
$\Leftrightarrow -\dfrac{25}{2} \leq -\dfrac{25}{2}\sin2x \leq \dfrac{25}{2}$
$\Leftrightarrow -\dfrac{1}{2} \leq 12 - \dfrac{25}{2}\sin2x \leq \dfrac{49}{2}$
Hay $-\dfrac{1}{2} \leq y \leq \dfrac{49}{2}$
Vậy $\min y = -\dfrac{1}{2} \Leftrightarrow \sin2x = 1 \Leftrightarrow x = \dfrac{\pi}{4} + k\pi$
$\max y = \dfrac{29}{2} \Leftrightarrow \sin2x = -1 \Leftrightarrow x = -\dfrac{\pi}{4} + k\pi \quad (k \in \Bbb Z)$
