f(x)=2sin^2x+cos(2x-π/3)-1
=(2sin^2x-1)+cos(2x-π/3)
=-cos2x+1/2*cos2x+√3/2*sin2x
=√3/2*sin2x -1/2*cos2x
=sin(2x-π/6)
∵x属于[π/12,π/2]
∴π/6≤2x≤π
即0≤2x-π/6≤5π/6
∴0≤sin(2x-π/6)≤1
即函数f(x)的值域为[0,1]