已知函数f(x)=2cosx(sinx cosx),x∈R 在△ABC中,f(A)=2,AB=1,AC=2,求BC

题目中缺了符号,应该是f(x)=2cosx(sinx+ cosx),
则f(x)=sin2x+2cos²x
=sin2x+cos2x+1
=√2*[sin2x*(√2/2)+cos2x*(√2/2)]+1
=√2*[sin2x*cos(π/4)+cos2x*sin(π/4)]+1
=√2sin(2x+π/4)+1
∵ f(A)=2
∴ √2sin(2A+π/4)+1=2
∴ sin(2A+π/4)=√2/2
∴ 2A+π/4=π/4或2A+π/4=3π/4
∴ A=0(舍)或A=π/4
利用余弦定理
BC²=AB²+AC²-2AB*AC*cosA
=1+4-2*1*2*（√2/2)
=5-2√2
∴ BC=√(5-2√2)