设函数F(x)=ab其中a=(2cox,1)b=(cosx,根号3sin2x),x属于R （1）若f(x)=1-根号3,

向量a=(2cosx,1),b=(cosx,根号3sin2x),
f(x)=向量a*向量b
=2cos²x+√3sin2x
=√3sin2x+cos2x+1
=2sin(2x+π/6)+1
∵f(x)=1-√3
∴2sin(2x+π/6)+1=1-√3
∴ sin(2x+π/6)=-√3/2
∵x∈[-π/3,π/3]
∴2x+π/6∈[-π/2,5π/6]
∴2x+π/6=-π/3
∴x=-π/4
(2)
函数y=2sin2x的图象按向量C=（m,n)
得到y-n=2sin[2(x-m)]
即y=2sin(2x-2m)+n
根据题意,y=2sin(2x-2m)+n
与f(x)=2sin(2x+π/6)+1为同一函数
∴n=1,2m=π/6+2kπ,k∈Z
∵|m|

f(x)=2cos^2 x+根号3sin2x
=1+cos2x+根号3sin2x
=2sin(2x+派/6)+1
1.
sin(2x+派/6)=-根号3/2
x=-派/12
2.
m=-派/12,n=1