6.(2023•咸阳模拟)已知方程sin2α+2sinαcosα−2sinα−4cosα=0\sin ^{2}\alpha +2\sin \alpha \cos \alpha -2\sin \alpha -4\cos \alpha =0sin2α+2sinαcosα−2sinα−4cosα=0,则cos2α−sinαcosα=(\cos ^{2}\alpha -\sin \alpha \cos \alpha =(cos2α−sinαcosα=( )))