(a)因式分解X^3-3X^2+3X-1 (b)利用(a)部的結果,解Sin^3 θ-3Sin^2 θ+3Sinθ=0,其中0°≦θ<360°? Sol.(a)(x-1)^3?
回答 (2)
2018-05-21 3:52 pm
✔ 最佳答案
(a)因式分解x^3－3x^2+3x－1Sol
x^3－3x^2+3x－1
=(x^3－x^2)+(－2x^2+2x)+(x－1)
=x^2(x－1)－2x(x－1)+(x－1)
=(x^2－2x+1)(x－1)^2
=(x－1)^3
(b)利用(a)部的結果,解Sin^3 θ－3Sin^2 θ+3Sinθ=0,其中0°≦θ<360°?
Sol
Sin^3 θ－3Sin^2 θ+3Sinθ=0
Sin^3 θ－3Sin^2 θ+3Sinθ－1=－1
(Sinθ－1)^3=－1
Sinθ－1=－1 (２虛根)
Sinθ=0
θ=0° or θ=180°
2018-05-23 11:37 am
(a) X³-3X²+3X-1
= (X³-2X²+X) -X²+2X-1
= X(X²-2X+1) - (X²-2X+1)
= (X-1)(X²-2X+1)
= (X-1)(X-1)(X-1)
= (X-1)³
(b) sin³θ - 3sin²θ + 3sinθ = 0
sin³θ - 3sin²θ + 3sinθ - 1 = -1
從(a): (sinθ - 1)³ = -1
sinθ-1 = -1
sinθ = 0
∴ θ = 0° 或 180° (0°≦θ<360°)
= (X³-2X²+X) -X²+2X-1
= X(X²-2X+1) - (X²-2X+1)
= (X-1)(X²-2X+1)
= (X-1)(X-1)(X-1)
= (X-1)³
(b) sin³θ - 3sin²θ + 3sinθ = 0
sin³θ - 3sin²θ + 3sinθ - 1 = -1
從(a): (sinθ - 1)³ = -1
sinθ-1 = -1
sinθ = 0
∴ θ = 0° 或 180° (0°≦θ<360°)
收錄日期: 2021-04-18 18:16:23
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