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
