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Veronichka1998
@Veronichka1998
August 2022
2
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Помогите пожалуйста! решите тригонометрическое уравнение:
2cos²x-11sin2x=12
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Лизаветк132
А ты в каком классе учишься???
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sofiya3009
Не знаю правильно это или нет
Решение
2cos²x - 11sin2x = 12
2cos²x - 22sinxcosx - 12cos²x - 12sin²x = 0
12sin²x + 22sinxcosx + 10cos²x = 0 делим на 2cos²x ≠ 0
6tg²x + 11tgx + 5 = 0
tgx = t
6t² + 11t + 5 = 0
D = 121 - 4*6*5 = 1
t₁ = (11 - 1)/12
t₁ = 5/12
t₂ = (11 + 1)/12
t₂ = 1
1) tgx = 5/12
x₁ = arctg(5/12) + πk, k ∈ Z
tgx = 1
x₂ = π/4 + πn, n ∈ Z
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Решение
2cos²x - 11sin2x = 12
2cos²x - 22sinxcosx - 12cos²x - 12sin²x = 0
12sin²x + 22sinxcosx + 10cos²x = 0 делим на 2cos²x ≠ 0
6tg²x + 11tgx + 5 = 0
tgx = t
6t² + 11t + 5 = 0
D = 121 - 4*6*5 = 1
t₁ = (11 - 1)/12
t₁ = 5/12
t₂ = (11 + 1)/12
t₂ = 1
1) tgx = 5/12
x₁ = arctg(5/12) + πk, k ∈ Z
tgx = 1
x₂ = π/4 + πn, n ∈ Z