Решение
2cos²x+4cos(pi/2-x)cos(x-4pi)+1=0
2cos²x + 4sinx*cosx + 1 = 0
2cos²x + 4sinx*cosx + sin²x + cos²x = 0
3cos²x + 4sinx*cosx + sin²x = 0 делим на cos²x ≠ 0
3 + 4tgx + tg²x = 0
tgx = t
t² + 4t + 3 = 0
t₁ = - 3
t₂ = - 1
tgx = - 3
x₁ = arctg(-3) + πn, n∈Z
x₁ = - arctg3 + πn, n∈Z
tgx = - 1
x₂= - π/4 + πk, k∈Z