f(x)=2√3sinxcosx+2（cosx）^2-1=√3sin2x+cos2x=2(sin2xcosπ/6+cos2xsinπ/6)=2sin(2x+π/6），2kπ+π/2≤2x+π/6≤2kπ+3π/2，单调递减，k∈Z，函数f(x)的单调递减区间：kπ+π/6≤x≤kπ+2π/3，k∈Z∴x∈[kπ+π/6，kπ+2π/3]，k∈Z。
