由2kπ-π/2≤2x-π/3≤2kπ+π/2,k∈Z
得2kπ-π/6≤2x≤2kπ+5π/6,k∈Z
即kπ-π/12≤x≤kπ+5π/12,k∈Z
所以函数递增区间为
[kπ-π/12,kπ+5π/12],k∈Z