是啊,我弄错了.
f（x)=√3sin2x-cos2x=2·[（√3/2）sin2x-（1/2）cos2x]=2·[sin2x·cos30°-sin30°·cos2x](据公式：sin(Φ-θ）=sinΦ·cosθ-sinθ·cosΦ）=2·sin(2x-30°)
又正弦函数的值域为[-1,1],所以2·sin(2x-30°)∈[-2,2],即f（x）∈[-2,2].