答：f(x)=6cos²x-2√3sinxcosx=3*(cos2x+1)-√3sin2x=3cos2x-√3sin2x+3=2√3*[(√3/2)cos2x-(1/2)sin2x]+3=2√3cos(2x+π/6)+31）f(x)的最小正周期T=2π/2=π值域为[3-2√3,3+2√3]2）f(B)=2√3cos(2B+π/6)+3...