已知f(x)等于sin(2x＋6)＋ cos(2x－3分之帕)求函数f(x) 的最小正周期

网友回答

f(x)=(1/2)sin2x+(√3/2)cos2x+(1/2)cos2x+(√3/2)sin2x
=[1+(√3)]/2(sin2x+cos2x)
=[1+(√3)]/√2(1/√2*sin2x+1/√2*cos2x)
=[1+(√3)]/√2sin(2x+π/4),∴T=π