f(x)=1+2√3sinxcosx+2cos^2x

f(x)=1+2√3sinxcosx+2cos^2x
=1+√3sin2x+2cos^2x
=1+2sin（2x+π/6)
所以f(x)的最小正周期T=2π/2=π

2kπ+π/2＜=2x+π/6＜=2kπ+3π/2
解得kπ+π/6＜=x＜=kπ+2π/3
所以f(x)的单调递减区间[kπ+π/6 , kπ+2π/3] k属于整数

解1.f(x)=1+2√3sinxcosx+2cos²x=1+√3sin2x+1+cos2x
=2(sin2xcosπ/6+cos2xsinπ/6)+2
=2sin(2x+π/6)+2
所以最小正周期T=2π/2=π
2.令2kπ+π/2≤2x+π/6≤2kπ+3π/2，k∈Z
解得kπ+π/6≤x≤kπ+2π/3,k∈Z
所以f(x)的单调递减区间为[kπ+π/6,kπ+2π/3],k∈Z