求y=sin2x+根号3cos2x的周期和值域

2025-06-25 09:48:46
推荐回答(1个)
回答1:

y=sin2x+√3*cos2x
=2(1/2*sin2x+√3/2*cos2x)
=2sin(2x+π/3)
函数的周期为:T=2kπ/2=kπ,k为整数
∵-1≤sin(2x+π/3)≤1
∴函数的值域为:[-2,2]