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1999mishkagammi
@1999mishkagammi
July 2022
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Помогите,пожалуйста. Нужно найти значение косинуса альфа и саму альфа.
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sharadi
Verified answer
5Cosα -2Sin²α -1 = 0
5Cosα - 2(1 - Cos²α) -1 = 0
5Cosα - 2 + 2Cos²α - 1 = 0
2Cos²α + 5Cosα -3 = 0
Cosα = t
2t² + 5t -3 = 0
D = b² -4ac = 25 -4*2*(-3)= 25 + 24 = 49>0 (2 корня)
t₁= (-5+7)/4 = 1/2; t₂= (-5 -7)/4 = -3
Cosα = 1/2
Cosα = -3
α = +-π/3 + 2πk , k ∈Z
∅
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Answers & Comments
Verified answer
5Cosα -2Sin²α -1 = 05Cosα - 2(1 - Cos²α) -1 = 0
5Cosα - 2 + 2Cos²α - 1 = 0
2Cos²α + 5Cosα -3 = 0
Cosα = t
2t² + 5t -3 = 0
D = b² -4ac = 25 -4*2*(-3)= 25 + 24 = 49>0 (2 корня)
t₁= (-5+7)/4 = 1/2; t₂= (-5 -7)/4 = -3
Cosα = 1/2 Cosα = -3
α = +-π/3 + 2πk , k ∈Z ∅