∴2cos2A+cosA- 1=0, the solution is obtained.

CosA= 1/2 or cosA=- 1.

And ∵ 0 < a < π, cosA=- 1 does not meet the meaning of the question, so it is discarded.

∴cosA= 1/2, available

a =π/3；

(ⅱ)∵A = 1，A=π/3，

According to cosine theorem a2=b2+c2-2bccosA,

You can get 1=b2+c2-2bccosπ/3.

=(b+c) 2-3 years BC,

And ∵b+c=2, ∴ 1=22-3bc, the solution is BC = 1.

So the area of △ABC is S△ABC√3.