解.令-π/2+2kπ≤2x-π/6≤π/2+2kπ,k∈Z
即-π/6+kπ≤x≤π/3+kπ,k∈Z
∵x∈[0,π]
∴令k=0则有-π/6≤x≤π/3
令k=1则有5π/6≤x≤4π/3
∴函数在[0,π]上的单调递增区间为[0,π/3]U[5π/6,π]