f(x)=(√3)sin(ωx+φ)-cos(ωx+φ)
=2{[(√3)/2]sin(ωx+φ)-(1/2)cos(ωx+φ)}
=2[sin(π/3)sin(ωx+φ)-cos(π/3)cos(ωx+φ)]
=-2cos(ωx+φ+π/3)
f(-x)=-2cos(-ωx+φ+π/3)
因?yàn)閒(-x)=f(x)
所以cos(φ+π/3-ωx)=cos(φ+π/3+ωx)
所以φ+π/3=kπ,k∈Z
因?yàn)?