（1）f﹙x﹚=向量a·向量b=2cos²x+2√3sinxcosx=√3sin2x+cos2x+1=2sin(2x+π/6)+1最小正周期T=2π／2=π由,2kπ-π/2≤2x+π/6≤2kπ+π/2,kπ-π/3≤x≤kπ+π/6,递增区间 [kπ-π/3 ,kπ+π/6],k∈Z.（2）将...2kπ-π/2≤2x+π/6≤2kπ+π/2,怎么来的令z=2x+π/6, 则f(x)=2sin(2x+π/6)+1=2sinz+1, y=sinz的递增区间是2kπ-π/2≤z≤2kπ+π/2, 还原，2kπ-π/2≤2x+π/6≤2kπ+π/2。