函數Y=2SIN[2X PAI/6]X是-PAI到0,單調減區間是__________詳解

熱心網友

函數y=2sin[2x+π/6],x∈[-π，0],單調減區間是？函數y=2sin[2x+π/6]的單調減區間為2kπ+π/2≤2x+π/6≤2kπ+3π/2kπ+π/6≤x≤kπ+2π/3 k∈Z因x∈[-π，0]所以k=-1y的單調減區間是[-5π/6，-π/3](剛才搞錯了。不知現在對否？）

熱心網友

-π=-11π/6=<2x+π/6=<π/6考慮基本函數y=sint,當t∈[-11π/6,-3π/2]或[-π/2,π/6]是增函數,在t∈[-3π/2,-π/2]時是減函數.解不等式:-11π/6=<2x+π/6=<-3π/2;-π/2=<2x+π/6=<π/6;以及-3π/2=<2x+π/6=<-π/2.得到-π=