∴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.