函数Y=2SIN[2X PAI/6]X是-PAI到0,单调减区间是__________详解
热心网友
函数y=2sin[2x+π/6],x∈[-π，0],单调减区间是？函数y=2sin[2x+π/6]的单调减区间为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的单调减区间是[-5π/6，-π/3](刚才搞错了。不知现在对否？）
热心网友
-π=-11π/6=<2x+π/6=<π/6考虑基本函数y=sint,当t∈[-11π/6,-3π/2]或[-π/2,π/6]是增函数,在t∈[-3π/2,-π/2]时是减函数.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=