已知函数f(x)=2根号3sinxcosx+2cos^2x 求函数最小正周期和值域

f(x)=2√3sinxcosx+2cos^2x
=√3*sin2x+cos2x+1
=2(√3/2*sin2x+1/2*cos2x)+1
=2sin(2x+π/6)+1
f(x)的最小正周期T=2π/2=π
f(x)min=-2+1=-1,f(x)max=2+1=3
f(x)值域为[-1,3]