f(x)=cos(2x-π/3)-(cos^2x-sin^2x) 　　=cos(2x-π/3)-cos2x 　　=2sin(2x-π/6)sinπ/6 　　=sin(2x-π/6) 　　因为y=sinx的单减区间为[π/2+2kπ,3π/2+2kπ](k为整数) 　　---->y=sin2x的单增区间为[π/4+kπ,3π/4+kπ](k为整数) 　　---->y=sin(2x-π/6)的单增区间为[π/6+kπ,2π/3+kπ] 　　---->y=-sin(2x-π/6)的单增区间为[π/6+kπ,2π/3+kπ] 　　f(a)=3/5---->sin(2a-π/6)=3/5 　　---->cos(2a-π/6)=4/5 　　sin(2a-π/6+π/6)=sin(2a-π/6)cosπ/6+cos(2a-π/6)sinπ/6 　　=3/5*√3/2+4/5*1/2 　　=(1/10)*(3√3+4) 　　即sin2a=(1/10)*(3√3+4)