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mrnobodyyyy
@mrnobodyyyy
July 2022
1
13
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Помогите, пожалуйста, с заданием 13 (тригонометрия, уравнение)
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nKrynka
Решение
cos2x - 2√2sin(π/2 + x) - 2 = 0
cos2x - 2√2cosx - 2 = 0
2cos²x - 1 -
2√2cosx - 2 = 0
2cos²x -
2√2cosx - 3 = 0
cosx = t
2t² - 2√2t - 3 = 0
D = (2√2)² + 4*2*3 = 8 + 24 = 32
t₁ = (2√2 - 4√2)/4 = -
√2/2
t
₂ = (2√2 + 4√2)/4 = 6√2/4 = 3√2/2 > 1, не удовлетворяет
условию: IcostI ≤ 1
cosx = - √2/2
x = (+ -) * arccos(- √2/2) + 2πk, k ∈ Z
x = (+ -)*(3
π/4
) + 2πk, k ∈ Z
2 votes
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Answers & Comments
cos2x - 2√2sin(π/2 + x) - 2 = 0
cos2x - 2√2cosx - 2 = 0
2cos²x - 1 - 2√2cosx - 2 = 0
2cos²x - 2√2cosx - 3 = 0
cosx = t
2t² - 2√2t - 3 = 0
D = (2√2)² + 4*2*3 = 8 + 24 = 32
t₁ = (2√2 - 4√2)/4 = - √2/2
t₂ = (2√2 + 4√2)/4 = 6√2/4 = 3√2/2 > 1, не удовлетворяет
условию: IcostI ≤ 1
cosx = - √2/2
x = (+ -) * arccos(- √2/2) + 2πk, k ∈ Z
x = (+ -)*(3π/4) + 2πk, k ∈ Z