Решите уравнение 2ctg^2 x+3ctgx-2=0

2ctg²(x) + 3ctg(x) - 2 = 0

• Пусть ctg(x) = t, тогда ctg²(x) = t²

• Получаем:

2t² + 3t - 2 = 0

(a = 2, b = 3, c = -2)

D = b² - 4ac

D = 3² - 4 • 2 • (-2) = 9 + 16 = 25 = 5²

t₁,₂ = (-b ± √D)/2a

t₁ = (-3 + 5)/2 • 2 = 2/4 = ½

t₂ = (-3 - 5)/2 • 2 = -8/4 = -2

• Получаем систему:

[ ctg(x) = ½

[ ctg(x) = -2

[ x₁ = arcctg(½) + πn, n ∈ ℤ

[ x₂ = arcctg(-2) + πn, n ∈ ℤ

[ x₁ = arcctg(½) + πn, n ∈ ℤ

[ x₂ = π - arcctg(2) + πn, n ∈ ℤ

x₁ = arcctg(½) + πn, n ∈ ℤ

x₂ = π - arcctg(2) + πn, n ∈ ℤ