题目

根号2sin(3x-π/4)的单增区间

提问时间：2021-03-13

答案

y=Asin(ax+b) (a>0,A>0)
递增区间[(2kπ)/a-π/(2a)-b/a,(2kπ)/a+π/(2a)-b/a]
递减区间[(2kπ)/a+π/(2a)-b/a,(2kπ)/a+3π/(2a)-b/a]
∴√2sin(3x-π/4)
递增区间[(2kπ)/3-π/12,(2kπ)/3,+π/4]
递减区间[(2kπ)/3,+π/4,(2kπ)/3+7π/12]