设函数f(x)=cos(2x+2π/3+2[(cosx)的平方],求f（x）的周期

设函数f(x)=cos(2x+2π/3+2[(cosx)的平方],求f（x）的周期

是f(x)=cos(2x+2π/3)+2cos²x吗?
如果是,
f(x)=cos2xcos2π/3-sin2xsin2π/3+2cos²x
=-cos2x/2-√3sin2x/2+1+cos2x
=cos2x/2-√3sin2x/2+1
=sin(2x+5π/6)+1
∴T=2π/2=π

f(x)=cos(2x+2π/3）+2[(cosx)的平方]
f(x)=cos(2x+2π/3）+2[(cosx)的平方]-1+1
f(x)=cos(2x+2π/3+cos2x+1
T=2π/ω
得出T=π