Question

已知函数f(x)=A sin(ωx+φ)(A>0,ω>0,-π/2<φ<π/2)的图像如图所示

求f(x)≥√2在区间[-π,π]上的解集

Answer 1

解析：∵函数f(x)=A sin(ωx+φ)(A>0,ω>0,-π/2<φ<π/2)
由图示所知：T/4=2π/3-π/6=π/2，∴T=2π，w=1；A=1
∴f(2π/3)=sin(2π/3+φ)=0==>2π/3+φ=π==>φ=π/3
∴f(x)=sin(x+π/3)
∵sin(x+π/3)>=√2，此处有问题函数最大值为1，怎么可能大于等于√2，无解

追问

不好意思我打错啦：求f(x)≥√2/2在区间[-π,π]上的解集？

追答

∵sin(x+π/3)>=√2/2
x+π/3=π/4==>x=-π/12
x+π/3=3π/4==>x=5π/12
f(x)≥√2/2在区间[-π,π]上的解集为-π/12<=x<=5π/12