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.