已知向量m=（2sinx/2,-根3）,n=（1-2sin²x/4,cosx）,其中x属于R
问题描述：
已知向量m=（2sinx/2,-根3）,n=（1-2sin²x/4,cosx）,其中x属于R
1、若m垂直于n,求x取值的集合
2、若f（x）=m*n-2t,当x属于【0,π】时函数f（x）有两个零点,求实数t的取值范围
答
(1)
m.n=0
(2sin(x/2),-√3).( 1- 2(sin(x/4))^2,cosx) =0
2sin(x/2).[ 1- 2(sin(x/4))^2] -√3cosx =0
2sin(x/2) cos(x/2)-√3cosx =0
sinx-√3cosx =0
tanx = √3
x = kπ+ π/3 k=0,1,2,.
(2)
f(x)=0
m.n -2t =0
sinx-√3cosx - 2t =0
2sin(x-π/3) -2t =0
t = sin(x-π/3)
0