已知函数f(x)=-根号3*sin^2 x+sinxcosx.求f(x)的最小正周期.
-√3*sin^2 x=-√3*【（1-cos2x）/2】=-√3/2+√3/2*cos2x
sinxcosx=1/2*sin2x
f(x)=-√3*sin^2 x+sinxcosx
=-√3/2+√3/2*cos2x+1/2*sin2x
=sin(2x+π/3)-√3/2
f(x)的最小正周期=2π/|ω|=π