Question

已知f(x)=sin(2wx-π/6)+1/2(w>0),最小正周期为π/2,写出f(x)的单调增区间,求f(x)单调增

Answer 1

最小正周期T=2π/(2w)=π/2,所以w=2

于是f(x)=sin(4x-π/6)+1/2

令2kπ-π/2≤4x-π/6≤2kπ+π/2

那么kπ/2-π/12≤x≤kπ/2+π/6

所以f(x)的单调递增区间为[kπ/2-π/12,kπ/2+π/6] (k∈Z)