$y = \sin^2x - \sin x\cos x - \cos^2x - 2$
$\to y + 2= - \dfrac{1}{2}\sin2x - \cos2x$
Phương trình có nghiệm
$\Leftrightarrow (y +2)^2 \leq \dfrac{1}{4} + 1$
$\Leftrightarrow (y +2)^2 \leq \dfrac{5}{4}$
$\Leftrightarrow -\dfrac{\sqrt5}{2} \leq y + 2 \leq \dfrac{\sqrt5}{2}$
$\Leftrightarrow -\dfrac{\sqrt5}{2} - 2 \leq y \leq \dfrac{\sqrt5}{2} - 2$
Vậy $\min y = -\dfrac{\sqrt5}{2} - 2$
$\max y = \dfrac{\sqrt5}{2} - 2$
