f(x)=向量a.向量b-1.
=√3sin2x+2cos^2x-1.
=√3sin2x+1+cos2x-1.
=2[(√3/2)sin2x+(1/2)cos2x].
∴f(x)=2sin(2x+π/6).
(1) 由 2kπ+π/2≤2x+π/6≤2kπ+3π/2,k∈Z.
得 kπ+π/6≤x≤kπ+2π/3,k∈Z.---即所求f(x)=2sin(2x+π/6)的单调递减区间；
(2) 由题设f(A)=2,得:2sin(2A+π/6)=2,
sin(2A+π/6)=1.
∴2A+π/6=π/2.
∴∠A=π/6.
由正弦定理,得：a/sinA=b/sinB.sinB=bsinA/a.
sinB=b*sin(π/6)/a.
=3*(1/2)/√3.
sinB=√3/2,
∴∠B=π/3,或∠B=2π/3.
则 ∠C=π/2,或∠C=π/6.
当C=π/2时,由勾股定理得：c=2√3；
当C=π/6时,则c=a=√3.
∴c=√3,或c=2√3.