问题描述: y=log底数是2 sin(-2x+三分之π)+1的单调增区间 1个回答 分类:数学 2014-10-17 问题解答: 我来补答 函数y=log(2) x是单调增函数所以函数y=log(2)[ sin(-2x+π/3)+1]函数的单调性由sin(-2x+π/3)+1决定函数y=sinx的单调区间是[2Kπ-π/2,2Kπ+π/2]单调递增[2Kπ+π/2,2Kπ+3π/2]单调递减2Kπ-π/2≤-2x+π/3≤2Kπ+π/22Kπ-5π/6≤-2x≤2Kπ+π/6-Kπ+5π/12≥x≥-Kπ-π/12所以Kπ-π/12≤x≤Kπ+5π/12时,函数单调递增同理可得Kπ+5π/12≤x≤Kπ+11π/12时,函数单调递减 展开全文阅读