Решите уравнение 2sinx^2 + (2- корень из 2)cos x +корень из двух -2=0

2sin²x + (2 - √2)cosx + √2 - 2 = 0 2 - 2cos²x + (2 - √2)cosx + √2 - 2 = 0 2cos²x + (√2 - 2)cosx - √2 = 0 cosx = (2 - √2 ± √(2 - 4√2 + 4 + 8√2))/4 = (2 - √2 ± √(√2 + 2)²)/4 = (2 - √2 ± (√2 + 2))/4 = {1; -√2/2} cosx = 1 => x = 2πn, n ∈ ℤ cosx = -√2/2 => x = π ± π/4 + 2πk, k ∈ ℤ Ответ: x = 2πn, n ∈ ℤ; x = π ± π/4 + 2πk, k ∈ ℤ