y=sinx單調(diào)增區(qū)間[2kπ-π/2,2kπ+π/2](k∈z)
所以sin(2x-π/6)的單調(diào)遞增區(qū)間為
2kπ-π/2≤2x-π/6≤2kπ+π/2,k∈Z
-π/6+kπ≤x≤π/3+kπ,k∈Z
令k=0則有-π/6≤x≤π/3
令k=1則有5π/6≤x≤4π/3
所以sin(2x-π/6)在[0,2π/3]上的單調(diào)遞增區(qū)間為[0,π/3]