三角函数问题:设函数y=√2sin(2ax+4π)的最小正周期为2π⼀3.

2025-06-22 16:38:39
推荐回答(1个)
回答1:

T=2π/2a=2π/3,a=3/2
y==√2sin(3x+4π),g(x)==√2sin[3(x-π/2)+4π]=√2sin(3x-3π/2)=√2sin(3x+π/2)=√2cos(3x)
-π+2kπ<=3x<=2kπ,-π/3+2/3kπ<=x<=2/3kπ,单增
2kπ<=3x<=π+2kπ,2/3kπ<=x<=π/3+2/3kπ,单减