f(x)=3√3sinxcosx-3cos²x+2
=(3√3/2)sin2x-3(1+cos2x)/2+2
=(3√3/2)sin2x-(3/2)cos2x+1/2
=√[(3√3/2)²+(3/2)²]sin(2x-z)+1/2
=3sin(2x-z)+1/2
其中tanz=(3/2)/(3√3/2)=√3/3
z=π/6
所以f(x)=3sin(2x-π/6)+1/2
所以T=2π/2=π
sin(2x-π/6)=1时有最大值
此时2x-π/6=2kπ+π/2
2x=2kπ+2π/3
所以当x∈{x|x=kπ+π/3,k∈Z}时,f(x)最大=3+1/2=7/2