f(x)=sin^22+2sinxcosx+3cos^2x
=1+2sinxcosx+2cos^2
=sin2x+2cos^2-1+1+1
=sin2x+cos2x+2
=√2(sin2x+π/4)+2
最小正周期:T=2π/2=π
單調(diào)遞增:
2x+π/4=-π/2+2kπ （k∈Z）
2x=-3π/4+2kπ
x=-3π/8+kπ
2x+π/4=π/2+2kπ （k∈Z）
2x=π/4+2kπ
x=π/8+kπ
單調(diào)遞增區(qū)間：[-3π/8,π/8]（k∈Z）