已知θ∈［0,2π),而sinθ、cosθ是方程x2-kx+k+1=0的两实数根,求k和θ的值.∵sinθ、cosθ是方程x2-kx+k+1=0的两实数根,∴代入(sinθ+cosθ)2=1+2sinθcosθ中整理,可得k2=1+2(k+1),即k2-2k-3=0.∴k=-1或k=3(舍).代回原方程组得∴或即θ=π或θ=.2分之3π 最后θ等于π和2分之3π不明白啊

(x-sinθ)*(x-cosθ) = x^2-kx+k+1=0
so sinθ+cosθ = k,sinθ*cosθ = k+1,so sin(θ+PI/4) = k/sqrt(2),sin(2θ) = 2(k+1),
if k=-1,so sin(θ+PI/4) = -1/sqrt(2),sin(2θ) = 0,so 2θ = 0,π,2π,3π,
so θ = 0,π/2,π,3π/2,考虑sin(θ+PI/4) = -1/sqrt(2),so θ = π,3π/2,