Question

已知向量m=(cosx/2,-1),n=(√3sinx/2,cos²x/2),设函数f(x)=m•

已知向量m=(cosx/2,-1),n=(√3sinx/2,cos²x/2),设函数f(x)=m•n，
（1）求函数f(x)的单调递增区间
（2）求函数f(x)在x∈[0,π]上的零点

Answer 1

(1) f(x)=m*n
=cos(x/2)*√3sin(x/2)+(-1)*cos^2(x/2)
=√3/2sinx-1/2(cosx+1)
=sin(x-π/6)-1/2
-π/2+2kπ=<x-π/6<=π/2+2kπ
-π/3=2kπ=<x<=2π/3+2kπ
单调递增区间：[-π/3+2kπ,2π/3+2kπ]
(2)
f(x)=0
sin(x-π/6)-1/2=0
sin(x-π/6)=1/2
x-π/6=π/6
x=π/3
或者x-π/6=5π/6
x=π
零点为：x=π/3或者x=π

Answer 2

如果有f(x)=m.n,则有：
f(x)=√3sin(x/2)cos(x/2)-cos^2(x/2)
=√3/2sinx-(1+cosx)/2
=√3/2sinx-(1/2)cosx-1/2
=sin(x-π/6)-1/2

(1)求f(x)的单调增区间，则有：
2kπ-π/2<=x-π/6<=2kπ+π/2
2kπ-π/2+π/6<=x<=2kπ+π/2+π/6
2kπ-π/3<=x<=2kπ+2π/3,k∈Z.
即单调增区间为：[2kπ-π/3,2kπ+2π/3].

(2)当f(x)=0,则有：
sin(x-π/6)=1/2
所以：
x-π/6=2kπ±π/6
x=2kπ或者x=2kπ+π/3,k∈Z.

Answer 3

f(x)=m•n=√3/2•sinx-1/2•cosx-1/2
=sin(x-π/6)-1/2
此题重点在于三角函数的化简，一般套路都是二倍角公式+辅助角公式
而后单调区间零点啥的多看看课本吧，楼上给出了详细的解答

Answer 4