已知f(x)=2sin(2x+π/6)+a+1,a为常数,求f(x)的递增区间.
-π/2+2k≤2x+π/6≤π/2+2kπ
-2π/3+2kπ≤2x≤π/3+2kπ
故有 -π/3+kπ≤x≤π/6+kπ
即单增区间为[-π/3+kπ,π/6+kπ] (k∈Z)