f(x)=2-cos(2x-π/3)-2sin²x
=2-1/2cos2x-√3/2sin2x-1+cos2x
=1+1/2cos2x-√3/2sin2x
=1+cos(2x+π/3)
因为x∈{0,π/2},则2x+π/3∈{π/3,4π/3},
所以f(x)∈{0,3/2}即：f(x)的值域为：{0,3/2}