f(x)=sin(x+π/3)的单调区间是？

sin(y)单调区间 增 2kπ<=y<=2kπ+π
减 2kπ-π<=y<=2kπ
sin(x+π/3) 中 y=x+π/3
所以增 2kπ<=x+π/3<=2kπ+π
减 2kπ-π<=x+π/3<=2kπ
即增 2kπ-π/3<=x<=2kπ+π-π/3
减 2kπ-π-π/3<=x<=2kπ-π/3

因为f(x)=sin(x)单调增区间为 2kπ<=x<=2kπ+π
单调减区间为 2kπ-π<=x<=2kπ
所以f(x)=sin(x+π/3）中
增 2kπ<=x+π/3<=2kπ+π
减 2kπ-π<=x+π/3<=2kπ
f(x)的单调增区间为2kπ-π/3<=x<=2kπ+π-π/3
f(x)的单调减区间为2kπ-π-π/3<=x<=2kπ-π/3

k为整数
2kπ-5π/6<x<2kπ+π/6 递增
2kπ+π/6<x<2kπ+7π/6 递减