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](/uploads/image/z/3743165-29-5.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fa%3D%28sinx%2C%E2%88%9A3cosx%29%2C%E5%90%91%E9%87%8Fb%3D%28cosx%2Ccosx%29%2Cf%28x%29%3D%E5%90%91%E9%87%8Fa%C3%97%E5%90%91%E9%87%8Fb%EF%BC%881%EF%BC%89%E8%8B%A5%E5%90%91%E9%87%8Fa%E2%8A%A5%E5%90%91%E9%87%8Fb%EF%BC%8C%E6%B1%82x%E7%9A%84%E8%A7%A3%E9%9B%86%EF%BC%9B%EF%BC%882%EF%BC%89%E6%B1%82f%28x%29%E7%9A%84%E5%91%A8%E6%9C%9F%E5%8F%8A%E5%A2%9E%E5%8C%BA%E9%97%B4)
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