已知点A(cosx,1+cos2x),B(-λ+√3*sinx,cosx),x∈(0,π),向量a=(1,0).(1)若向量BA与a共线,求实数的值,(2)若向量BA⊥a,求实数λ的取值范围.

BA=(cosx-(√3)sinx+λ,1+cos2x-cosx)
(1)若BA与a共线,则1+cos2x-cosx=0,在结合x∈(0,π)
解得:x=π/3,或x=π/2.
(2)若BA⊥a,则cosx-(√3)sinx+λ=0
==> λ=(√3)sinx-cosx=2sin(x-π/6)
x∈(0,π) ==> x-π/6∈(-π/6,5π/6) ==> sin(x-π/6)∈(-1/2,1)
==> λ∈(-1,2).