Решение
√[2*sin(x + 3x) * (2cos3x*cos(3(π/2 + x))) + 4] /2 = √2*√2
√[2*sin(4x) * (2cos3x*cos(3π/2 + 3x)) + 4] /2 = 2
√[2*sin(4x) * (2cos3x*sin3x) + 4] /2 = 2
√[2*sin(4x) * sin(6x) + 4] /2 = 2
√[sin(4x) * sin(6x) + 2] = 2
√[sin(4x) * sin(6x)] = 0
{√[sin(4x) * sin(6x)]}² = 0
sin(4x) * sin(6x) = 0
1) sin4x = 0
4x = πk, k ∈Z
x₁ = πk/4, k∈Z
2) sin6x = 0
6x = πn, n ∈Z
x₂ = πn/6, n∈Z
Ответ: x₁ = πk/4, k∈Z ; x₂ = πn/6, n∈Z