紧急求教：向量a=(2cosx,1),b=(cosx,根号3sin2x-1),设函数f(x)=向量a*向量b，其中x∈R

求函数的最小正周期和单调增区间。

解答：
f(x)=向量a.向量b
=(2cosx,1).(cosx,√3sin2x-1)
=2cos²x+√3sin2x-1
=√3sin2x+cos2x
=2[sin(2x)*(√3/2)+cos(2x)*(1/2)]
=2[sin(2x)cos(π/6)+cos(2x)sin(π/6)]
=2sin(2x+π/6)
(1) T=2π/2=π
(2) 2kπ-π/2≤2x+π/6≤2kπ+π/2
2kπ-2π/3≤2x≤2kπ+π/3
kπ-π/3≤x≤kπ+π/6
所以，增区间为【kπ-π/3, kπ+π/6】,k∈Z

(1)f(x)=2cos x+ sin2x=1+2sin(2x+ ∴f(x)的最小正周期为π. (2)∵f(a)=2,即1+2sin(2a＋π/6)=2, ∴sin(2a+π/6) =1/2 ∵ π/6＜2a+ ＜13π/6 ∴2a+π/6 = 5π/6. 由cosa=1/2 =(b^2+c^2-a^2)/2bc 即(b+c)^2 -a^2 =3bc, ∴bc=2.又b+c=3(b＞c),∴b=2,c=1