f(x)＝√3sin2x+2sin²x＝√3sin2x-(1-2sin²x)+1＝√3sin2x-cos2x+1＝2[sin2xcos(π/6)-cos2xsin(π/6)]+1＝2sin(2x-π/6)+1.∵-1≤sin(2x-π/6)≤1,∴sin(2x-π/6)＝-1时，f(x)|min＝-1.sin(2x-π/6)＝1时，f(x)|max＝3。
