f(x)=√3sin＾2x+sinxcosx-√3/2=-√3/2*cos2x+1/2*sin2x=sin(2x-π/3),f(α)=sin(2α-π/3)=4/5,α∈（0,π/2）,2α-π/3∈（-π/3,2π/3）,∴cos(2α-π/3)=土3/5,∴cos2α=cos(2α-π/3+π/3)=土3/5*1/2-4/5*√3/...