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