解.sinα-2cosα=0即tanα=2
∴sin2α=2tanα/(1+tan²α)=4/5
∴1/(sinαcosα)=2/sin2α=5/2
∴cos2α=(1-tan²α)/(1+tan²α)=(1-4)/(1+4)=-3/5
4sin²α-3sinαcosα-5cos²α=2(1-cos2α)+(3/2)sin2α-(5/2)(1+cos2α)
=(3/2)sin2α-(9/2)cos2α-1/2
=(3/2)*(4/5)-(9/2)*(-3/5)-1/2
=17/5