$y=(\sin x-\cos x)^2 +2\cos 2x +\dfrac{3}{2}\sin 2x$
$=1-\sin 2x +2\cos 2x +\dfrac{3}{2}\sin 2x$
$=\dfrac{1}{2}\sin 2x +2\cos 2x +1$
Hay $\dfrac{1}{2}\sin 2x +2\cos 2x +1-y=0$
phương trình có nghiệm khi $(\dfrac{1}{2})^2+2^2 \ge (1-y)^2$
$\Leftrightarrow -y^2 +2y +\dfrac{13}{4}\ge0$
$\Leftrightarrow \dfrac{1}{2}(2-\sqrt{17}) \le y \le \dfrac{1}{2}(2+\sqrt{17})$ XONG NHÉ