求y=sin(π/3-x/2)在〔-2π,2π〕的单调递增区间
问题描述:
求y=sin(π/3-x/2)在〔-2π,2π〕的单调递增区间
答
y=sinx的单调增区间为(-π/2+2kπ,π/2+2kπ)
所以y=sin(π/3-x/2)单调增区间为
-π/2+2kπ<π/3-x/2<π/2+2kπ
解得π/3+4kπ<x<-5π/3+4kπ