Ответ Автор - nKrynka Ctg(π/4 - 2x) = √3- Ctg(2x - π/4) = √3 Ctg(2x - π/4 ) = - √32x - π/4 = arcctg(-√3) + πn, n∈Z2x - π/4 = 5π/6 + πn, n∈Z2x = 5π/6 + π/6 + πn, n∈Z2x = π + πn, n∈Zx = π/2 + (πn)/2 n∈Z