问题描述: y=sin((1/2)x+(π/3))的单调递增区间 1个回答 分类:数学 2014-11-04 问题解答: 我来补答 y=-sin(x/2-π/3)所以y增区间即sin(x/2-π/3)的减区间sinx的减区间是(2kπ+π/2,2kπ+3π/2)2kπ+π/2<x/2-π/3<2kπ+3π/22kπ+5π/6<x/2<2kπ+11π/64kπ+5π/3<x<4kπ+11π/3k=-1,则-7π/3<x<-π/3和[-2π,2π]交集是[-2π,-π/3)k=0,则5π/3<x<11π/3和[-2π,2π]交集是(5π/3,2π]所以y的增区间是(-2π,-π/3)∪(5π/3,2π) 展开全文阅读