=cos(2x+π/6)+sin2x=cos2xcosπ/6-sin2xsinπ/6+sin2x=(√3/2)cos2x-(1/2)sin2x+sin2x=(√3/2)cos2x+(1/2)sin2x=cos2xcosπ/6+sin2xsinπ/6=cos(2x-π/6)所以,f（x）最小正周期为：π第一步怎样的出来的？二倍角公式呀cos2a = 2cos²a - 1所以2cos²a = cos2a + 1