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Quartz51
@Quartz51
July 2022
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Пожалуйста, помо гите с уровнением 6sin^2x + 5cosx -5 = 0
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flsh
6·sin²x + 5·cos x - 5 = 0
6·(1 - cos²x) + 5·cos x - 5 = 0
6 - 6·cos²x + 5·cos x - 5 = 0
- 6·cos²x + 5·cos x + 1 = 0
6·cos²x - 5·cos x - 1 = 0
cos x = t
6t² - 5t -1 = 0
D = 25 + 24 = 49
t = (5 (+/-) 7)/12
t₁ = -1/6
t₂ = 1
cos x = -1/6
cos x = 1
x = (+/-) arc cos (-1/6) + 2πn, n ∈ Z
x = 2πk, k ∈ Z
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Answers & Comments
6·(1 - cos²x) + 5·cos x - 5 = 0
6 - 6·cos²x + 5·cos x - 5 = 0
- 6·cos²x + 5·cos x + 1 = 0
6·cos²x - 5·cos x - 1 = 0
cos x = t
6t² - 5t -1 = 0
D = 25 + 24 = 49
t = (5 (+/-) 7)/12
t₁ = -1/6
t₂ = 1
cos x = -1/6
cos x = 1
x = (+/-) arc cos (-1/6) + 2πn, n ∈ Z
x = 2πk, k ∈ Z