已知函数f(x)=sin^2*x+更号3*sinx*cosx+2cos^2*x,x属于R.(1)求函数f(x)的最小正周期和单调递增区间

f(x)=(sin²x+cos²x)+√3sinxcosx+cos²x=1+√3/2*sin2x+(1+cos2x)/2=√3/2sin2x+1/2*cos2x+3/2=sin2xcosπ/6+cos2xsinπ/6+3/2=sin(2x+π/6)+3/2所以T=2π/2=πsin(2x+π/6)递增则2kπ-π/2