求函数f（x）=cos2x-sin2x+2根号3sinxcosx的最小周期、最大值

f(x)＝cos2x-sin2x+2√3sinxcosx＝cos2x-sin2x+√3sin2x＝cos2x+(-1+√3)sin2x＝√(5-2√3)sin(2x+φ)＝(√5-√3)si(2x+φ).故最小正周期T＝2π/2＝π；最大值：f(x)|max＝√5-√3;最小值：f(x)|min＝√3-√5....