cos(3x)•cos(π/4) + sin(3x)•sin(π/4) = - 0,5

cosα•cosβ + sinα•sinβ = cos(α - β)

cos(3x - (π/4)) = - 1/2

[ 3x - (π/4) = (2π/3) + 2πn ⇔ 3x = (11π/12) + 2πn ⇔ x = (11π/36) + (2πn/3) , n ∈ Z

[ 3x - (π/4) = (-2π/3) + 2πk ⇔ 3x = (-5π/12) + 2πk ⇔ x = (-5π/36) + (2πk/3) , k ∈ Z

ОТВЕТ: (11π/36) + (2πn/3), n ∈ Z ; (-5π/36) + (2πk/3), k ∈ Z

Verified answer

2 строчку получаем из формулы разности аргументов, в данном случае собираем по формуле:

cos(x-y) =cosXcosY+sinXsinY