Question

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的值
(2)求f（x）的周期及递增区间。

Answer 1

(1)a⊥b
则：f(x)=sinxcosx+√3cosxcosx=sin2x/2+√3(1+cos2x)/2
=sin(2x+π/3)+√3/2
=0
∴2x+π/3=2kπ+3π/2±π/6
∴x=kπ+7π/12±π/12
(2)f(x)=sin(2x+π/3)+√3/2
∴f(x)周期为π，递增区间为[kπ-5π/12,kπ+π/12]