$y = \dfrac{\cos^2x + \sin x\cos x}{1 + \sin^2x}$
$\to y = \dfrac{1 + \cos2x + \sin2x}{3 - \cos2x}$
$\to 3y - y\cos2x = 1 + \cos2x + \sin2x$
$\to \sin2x + (y+1)\cos2x = 3y - 1$
Phương trình có nghiệm
$\Leftrightarrow 1^2 + (y+1)^2 \geq (3y - 1)^2$
$\Leftrightarrow 8y^2 - 8y - 1 \leq 0$
$\Leftrightarrow \dfrac{2 - \sqrt6}{4} \leq y \leq \dfrac{2 + \sqrt6}{4}$
Vậy $\min y = \dfrac{2 - \sqrt6}{4} ; \, \max y = \dfrac{2 + \sqrt6}{4}$
