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