作业帮已知向量a=(sinx,√3cosx),向量b=(cosx,cosx),f(x)=向量a×向量b（1）若向量a⊥向量b，求x的解集；（2）求f(x)的周期及增区间
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已知向量a=(sinx,√3cosx),向量b=(cosx,cosx),f(x)=向量a×向量b（1）若向量a⊥向量b，求x的解集；（2）求f(x)的周期及增区间
已知向量a=(sinx,√3cosx),向量b=(cosx,cosx),f(x)=向量a×向量b
（1）若向量a⊥向量b，求x的解集；（2）求f(x)的周期及增区间
已知向量a=(sinx,√3cosx),向量b=(cosx,cosx),f(x)=向量a×向量b（1）若向量a⊥向量b，求x的解集；（2）求f(x)的周期及增区间
f(x)=向量a×向量b=(sinx,√3cosx)*(cosx,cosx)
=sinxcosx+√3cosxcosx
=1/2(2sinxcosx+2√3cosxcosx)
=1/2(sin2x+√3cos2x+√3)
=1/2sin2x+√3/2cos2x+√3/2
=cosπ/3sin2x+sinπ/3cos2x+√3/2
=sin(2x+π/3)+√3/2
(1)sin(2x+π/3)+√3/2 =0
sin(2x+π/3)=-√3/2
2x+π/3=2kπ-π/3
x=kπ-π/3
(2)T=2π/2=π
由2kπ-π/2≤2x+π/3≤2kπ+π/2
解得增区间为：kπ-5π/12≤x≤kπ+π/12
f(x)=sinxcosx+√3(cosx)^2
=sin2x/2+√3(1+cos2x)/2
=sin(2x+π/3)+√3/2
1.向量a⊥向量b
则sin(2x+π/3)+√3/2=0即sin(2x+π/3)=-√3/2
2x+π/3=2kπ-π/3 ===>kπ-π/3
2.T=2π/2=π
2kπ-π/2<=2x+π/3<=2kπ+π/2
kπ-5π/12<=x<=kπ+π/12
f(x)=向量a×向量b
=sinxcosx+√3cos²x
=(1/2)sin2x+(√3/2)(cos2x+1)
=(1/2)sin2x+(√3/2)cos2x+√3/2
=sin(2x+π/6)+√3/2=0
sin(2x+π/6)=-√3/2
所以2x+π/6=2kπ-π/3或2kπ+4π/3
x=kπ-π/4或kπ+7π/12 k∈Z