设函数f(x)=2cos^2x+2根号3sinx*cosx,求f(x)的最小正周期以及单调增区间

f(x)=2cos²x+2√3sinxcosx=(2cos²x-1)+1+√3×2sinxcosx=cos2x+√3sin2x+1=2sin(2x+π/6)+1函数f(x)最小正周期T=2π/2=π由2kπ≤2x+π/6≤2kπ+π/2 (k∈z),得kπ-π/12≤x≤kπ+π/6或 由2kπ+3π/2≤2x...