解:2k1π<α<2k1π+,2k2π+π<β<2k2π+,k1、k2∈Z,
2(k1+k2)π+π<α+β<2(k1+k2)π+2π.
令k1+k2=n∈Z,
∴2nπ+π<α+β<2nπ+2π,n∈Z.