f(x)=2sin(2x+π/6),求g(x)=f(x-π/12)-f(x+π/12)的单调递增区间
问题描述：
f(x)=2sin(2x+π/6),求g(x)=f(x-π/12)-f(x+π/12)的单调递增区间
答
g(x)=f(x-π/12)-f(x+π/12)
=2sin(2x)-2sin(2x+π/3)
=2[sin(2x)-sin(2x+π/3)]
=2sin(2x-π/3)
单调递增区间由
2kπ-π/2≤2x-π/3≤2kπ+π/2
得kπ-π/12≤x≤kπ+5π/12