已知函数f(x)＝2cos^2(ωx/2)+cos(ωx+π/3)(ω＞0）

f(x)＝2cos^2(ωx/2)+cos(ωx+π/3)
=1+cos(ωx)+cos(ωx+π/3)
=1+cos(ωx)+cos(ωx)cosπ/3-sin(ωx)sinπ/3
=1+3cos(ωx)/2-√3sin(ωx)/2
=1+√3*[√3/2cos(ωx)-1/2sin(ωx)]
=1+√3*[sinπ/3cos(ωx)-cosπ/3sin(ωx)]
=1+√3*sin(π/3-ωx)
或者直接用和差化积
f(x)＝2cos^2(ωx/2)+cos(ωx+π/3)
=1+cos(ωx)+cos(ωx+π/3)
=1+2cos[(ωx+ωx+π/3)/2]cos(ωx-ωx-π/3)/2]
=1+2cos(ωx+π/6)cos(π/6)
=1+√3cos(ωx+π/6)

f(x)＝2cos^2(ωx/2)+cos(ωx+π/3)
=2cos^2(ωx/2)-1+cos(ωx+π/3)+1
=cos(ωx)+cos(ωx+π/3)+1
=cos(ωx)+cos(ωx)cos(π/3)+sin(ωx)sin(π/3)+1
=cos(ωx)+1/2cos(ωx)+√3/2sin(ωx)+1
=3/2cos(ωx)+√3/2sin(ωx)+1
=√3[√3/2cos(ωx)+1/2sin(ωx)]+1
=√3cos(ωx+π/6)+1

若f(x)=2cos^2[(ωx)/2]+cos[ωx+(Π/3)](ω>0)
则f(x)=cosωx+1+(1/2)cosωx-(√3/2)sinωx
=(3/2)cosωx-(√3/2)sinωx+1
=√3[(√3/2)cosωx-(1/2)sinωx]+1
=√3[cosωxcosπ/6-sixωxsinπ/6]+1
=√3cos(ωx+π/6)+1
因为f(x)的最小正周期为π，
所以ω=2π/π=2。