设f(x)=cos的平方+2/根号3sin2x.求f(x)的最小正周数

f(x)=cos^2x+√3/2sin2x
=1/2(cos2x+1)+√3/2sin2x
=√3/2sin2x+1/2cos2x+1/2
=sin2xcosπ/6+cos2xsinπ/6+1/2
=sin(2x+π/6)+1/2
所以最小正周期T=2π/2=π