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polyarny
@polyarny
July 2022
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Решите пожалуйста:
шарага
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Nymeria
2sin²x + sinx - 3 = 0
пусть sinx = a, -1 ≤ a ≤ 1
2a² + a - 3 = 0
D = 1² - 4*2*(-3) = 25
a₁ = (-1 + 5) / 4 = 1
a₂ = (-1 - 5) / 4 = -1,5 - не принадлежит ОДЗ
sinx = 1
x = π/2 + 2πk, k ∈ Z
tg²x + tgx - 2 = 0
пусть tgx = a, -1 ≤ a ≤ 1
a² + a - 2 = 0
D = 1² + 4*(-2) = 9
a₁ = (-1 + 3) / 2 = 1
a₂ = (-1 - 3) / 2 = -2 - не принадлежит ОДЗ
tgx = 1
x = π/4 + πk, k ∈ Z
2sin²x + 3cosx = 0
2(1 - cos²x) + 3cosx = 0
-2cos²x + 3cosx + 2 = 0 |*(-1)
2cos²x - 3cosx - 2 = 0
пусть cosx = a, -1 ≤ a ≤ 1
2a² - 3a - 2 = 0
D = (-3)² - 4*2*(-2) = 25
a₁ = (3 + 5) / 4 = 2 - не принадлежит ОДЗ
a₂ = (3 - 5) / 4 = -0,5
cosx = -0,5
x =
±3
π
/2 + 2
π
k, k
∈ Z
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polyarny
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Answers & Comments
пусть sinx = a, -1 ≤ a ≤ 1
2a² + a - 3 = 0
D = 1² - 4*2*(-3) = 25
a₁ = (-1 + 5) / 4 = 1
a₂ = (-1 - 5) / 4 = -1,5 - не принадлежит ОДЗ
sinx = 1
x = π/2 + 2πk, k ∈ Z
tg²x + tgx - 2 = 0
пусть tgx = a, -1 ≤ a ≤ 1
a² + a - 2 = 0
D = 1² + 4*(-2) = 9
a₁ = (-1 + 3) / 2 = 1
a₂ = (-1 - 3) / 2 = -2 - не принадлежит ОДЗ
tgx = 1
x = π/4 + πk, k ∈ Z
2sin²x + 3cosx = 0
2(1 - cos²x) + 3cosx = 0
-2cos²x + 3cosx + 2 = 0 |*(-1)
2cos²x - 3cosx - 2 = 0
пусть cosx = a, -1 ≤ a ≤ 1
2a² - 3a - 2 = 0
D = (-3)² - 4*2*(-2) = 25
a₁ = (3 + 5) / 4 = 2 - не принадлежит ОДЗ
a₂ = (3 - 5) / 4 = -0,5
cosx = -0,5
x = ±3π/2 + 2πk, k ∈ Z