Решение
2 – 2cos²(x) + (√2)sin(x) = 0
2 – 2*(1 - sin²x) + (√2)sinx = 0
2 – 2 + 2*sin²x + (√2)sinx = 0
2*sin²x + (√2)sinx = 0
sinx(2sinx + √2) = 0
1) sinx = 0
x₁ = πk, k ∈ Z
2) 2sinx + √2 = 0
sinx = - √2/2
x = (-1)^n * arcsin(- √2/2) + πn, n ∈ Z
x₂ = (-1)^(n + 1) * (π/4) + πn, n ∈ Z