Решение
2sin²x - 5cosx - 5 = 0
2*(1 - cos²x) - 5cosx - 5 = 0
2 - 2cos²x - 5cosx - 5 = 0
2cos²x + 5cosx + 3 = 0
cosx = t, I t I ≤ 1
2t² + 5t + 3 = 0
D = 25 - 4*2*3 = 1
t₁ = (- 5 - 1)/4 = - 6/4 = - 1,5 не удовлетворяет условию I t I ≤ 1
t₂ = (- 5 + 1)/4 = - 1
cosx = - 1
x = π + 2πk, k ∈ Z