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
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![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](/uploads/image/z/986124-12-4.jpg?t=%E5%B7%B2%E7%9F%A5%CE%B8%E2%88%88%EF%BC%BB0%2C2%CF%80%29%2C%E8%80%8Csin%CE%B8%E3%80%81cos%CE%B8%E6%98%AF%E6%96%B9%E7%A8%8Bx2-kx%2Bk%2B1%3D0%E7%9A%84%E4%B8%A4%E5%AE%9E%E6%95%B0%E6%A0%B9%2C%E6%B1%82k%E5%92%8C%CE%B8%E7%9A%84%E5%80%BC.%E2%88%B5sin%CE%B8%E3%80%81cos%CE%B8%E6%98%AF%E6%96%B9%E7%A8%8Bx2-kx%2Bk%2B1%3D0%E7%9A%84%E4%B8%A4%E5%AE%9E%E6%95%B0%E6%A0%B9%2C%E2%88%B4%E4%BB%A3%E5%85%A5%28sin%CE%B8%2Bcos%CE%B8%292%3D1%2B2sin%CE%B8cos%CE%B8%E4%B8%AD%E6%95%B4%E7%90%86%2C%E5%8F%AF%E5%BE%97k2%3D1%2B2%28k%2B1%29%2C%E5%8D%B3k2-2k-3%3D0.%E2%88%B4k%3D-1%E6%88%96k%3D3%28%E8%88%8D%29.%E4%BB%A3%E5%9B%9E%E5%8E%9F%E6%96%B9)
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
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
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
(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,