2сos²x + cosx - 1 = 0
Пусть t = cosx, t ∈ [-1; 1]
2t² + t - 1 = 0
D = 1 + 8 = 9 = 3²
t₁ = (-1 + 3)/4 = -1/2
t₂ = (-1 - 3)/4 = -1
Обратная замена:
cosx = -1/2
x = ±2π/3 + 2πn, n ∈ Z
cosx = -1
x = π + 2πk, k ∈ Z.
Ответ: x = ±2π/3 + 2πn, n ∈ ℤ; π + 2πk, k ∈ Z.