已知f(x)=sin^2x+2sinxcosx+3cos^2x,x∈(0,兀),求函数的最小正周期和值域;求函数的单调递增区间

2cos²x-1=cos2xf(x)=sin²x+2sinxcosx+3cos²x=(sin²x+cos²x)+(2cos²x-1)+1+2sinxcosx=2+sin2x+cos2x=2+(√2)sin[2x+(π/4)]T=2π/ω=πx∈(0,π)[2x+(π/4)]∈(π/4,9π/4)f(x)max=2+(...