若函数y=sin(kx/3+π/3)有一条对称轴方程为x=π/6,则k的值是答案是18T+3

问题描述:

若函数y=sin(kx/3+π/3)有一条对称轴方程为x=π/6,则k的值是
答案是18T+3

若函数y=sin(kx/3+π/3)有一条对称轴方程为x=π/6
那么x=π/6时,y=1或-1
kπ/18+π/3=π/2+2πT或kπ/18+π/3=3π/2+2πT
kπ/18+π/3=π/2+2πT
k=36T+3=18*2T+3
kπ/18+π/3=3π/2+2πT
k=36T+21=36T+18+3=18*(2T+1)+3
两个解集的并集
k=18T+3

我们知道sinx的对称轴为π/2+Tπ
所以把x=π/6带入(k/3+π/3=(π/2+Tπ)得k=18T+3

y的对称轴就是Y取最值的地方,(对三角函数就是这样)
当kx/3+π/3=π/2+nπ(n为整数)时,y取最值
由题意,x=π/6为该方程一解
代入得,
kπ/18+π/3=π/2+nπ,
k/18+1/3=1/2+n,
k/18=1/6+n,
k=3+18n,(n为整数)

由题知sin(k*π/18+π/3)=1或-1,k*π/18+π/3=(2n+1)π/2,k=18n+3,n为整数

kx/3+派/3=派/2+k派 把x=派/6代入求得 k=-3/17

kx/3+π/3=kπ+π/2

当x=π/6时,y=sin(kx/3+π/3)=±1,
所以kx/3+π/3终边落在y轴上
将x=π/6代入kx/3+π/3,得:kπ/18+π/3=π/2+Tπ(T∈Z)
所以k=18T+3(T∈Z)

sinx对称轴就是sinx取最值的地方
即x=kπ+π/2
所以这里kx/3+π/3=kπ+π/2
kx/3=kπ+π/6
x=3π+π/(2k)
x=π/6
3π+π/(2k)=π/6
3+1/(2k)=1/6
1/(2k)=-17/6
k=-3/17

函数y=sin(kx/3+π/3)对称轴方程:
(nπ+π/2-π/3)/(k/3)=π/6,
k=18n+3 n∈Z