求三角函数y=sin(π/3-2x)的单调递增区间
问题描述：
求三角函数y=sin(π/3-2x)的单调递增区间
答
y=sin(π/3-2x)=-sin(2x-π/3)
2kπ+π/2≤2x-π/3≤2kπ+3π/2
kπ+5π/12≤x≤kπ+11π/12
故增区间是[kπ+5π/12,kπ+11π/12]