求函数y=√2sin(2x-π/4)的单调递减区间
问题描述：
求函数y=√2sin(2x-π/4)的单调递减区间
答
y=-2sin(2x-π/4)
y递减则sin(2x-π/4)递增
sinx增区间是(2kπ-π/2,2kπ+π/2)
则2kπ-π/22kπ-π/4kπ-π/8
答
π/2+2kπ≤2x-π/4≤3π/2+2kπ
3π/8+kπ≤x≤7π/8+kπ
答
y=2sin(2x+π/4)+1 y=sinx单调递减区间是 【2kπ-π/2,2kπ+π/2】所以2kπ-π/2