设函数f(x)=acos^2(ωx)-(根号3)asin(ωx)cos(ωx)+b的最小正周期为π(a=/=0,ω>0)

f(x)=acos^2(ωx)-sqrt(3)asin(ωx)cos(ωx)+b
=a*(1+cos(2ωx))/2-sqrt(3)*a*sin(2ωx)/2+b
=a*[ cos(2ωx))*1/2-sin(2ωx)*sqrt(3)/2 ] +a/2+b
=a*[ cos(2ωx))*cos(π/3)-sin(2ωx)*sin(π/3) ] +a/2+b
=a*cos(2ωx+π/3)+a/2+b
(1)由于f(x)的最小正周期为π且ω>0,得ω=1.
(2)f(x)的定义域为[-π/3,π/6],意味着-π/3