Question

已知函数f(x)=Asin(ωx+φ),x∈R(之中A>0,ω>0,0<φ<π/2)的周期为π,且图像上一个最低点为M(2π/3,-2)

(1)求f（x）的解析式

Answer 1

∵周期为π
且T=2π/ω=π
ω=2
∴f(x)=Asin(2x+φ)
∵最低点为M(2π/3,-2)
∴A=2
sin(2*2π/3+φ)=sin(4π/3+φ)=-1
当x=2π/3时，有最小值
即有2*2π/3+φ=2kπ-π/2 4π/3+φ=2kπ-π/2
φ=2kπ-π/2-4π/3=2kπ-11π/6=2kπ-2π+π/6=2(k-1)π+π/6
∵0＜φ＜π/2
∴φ=π/6
∴f(x)=2sin(2x+π/6)

Answer 2

A=2
π=2π/ω,
ω=2
2(2π/3)+φ=3π/2
φ=π/6
f(x)=2sin(2x+π/6),