已知f(x)=2sinxcosx-2根号3(cosα)^2+根号3,若f(α)=10/13,且α属于【π/2,π】,求s

f(x)=2sinxcosx-2√3(cosα)^2+√3
=sin2x-√3(2(cosα)^2-1)
=sin2x-√3cos2x
=2sin(2x-π/3)
f(α)=10/13
∴2sin(2α-π/3)=10/13
sin(2α-π/3)=5/13,∵α属于[π/2,π],2α-π/3为第二象限角
∴cos(2α-π/3)=-12/13
sin2α=sin(2α-π/3+π/3)
=sin(2α-π/3)*cosπ/3+cos(2α-π/3)*sinπ/3
=(5/13)*(1/2)+(-12/13)*(√3/2)
=(5-12√3)/26.