求sin(2x-π/6)在[0,2π/3]上的单调递增区间
问题描述：
求sin(2x-π/6)在[0,2π/3]上的单调递增区间
答
y=sinx单调增区间[2kπ-π/2,2kπ+π/2](k∈z)
所以sin(2x-π/6)的单调递增区间为
2kπ-π/2≤2x-π/6≤2kπ+π/2,k∈Z
-π/6+kπ≤x≤π/3+kπ,k∈Z
令k=0则有-π/6≤x≤π/3
令k=1则有5π/6≤x≤4π/3
所以sin(2x-π/6)在[0,2π/3]上的单调递增区间为[0,π/3]