За 90 ! решить уравнение сtg(3x-π/6)=√3 и найти все корни этого уравнения, принадлежащие отрезку [-5π/2; -2π].

сtg(3x - π/6) = √3
3x - π/6 = arcctg√3 + πn, n ∈ Z
3x - π/6 = π/6 + πn, n ∈ Z
3x = 2π/6 + πn, n ∈ z
3x = π/3 + πn, n ∈ Z
x = π/9 + πn/3, n ∈ Z

Отберем корни, принадлежащие отрезку [-5π/2; -2π].
при n = 0 х= π/9 ∉ [-5π/2; -2π],
при n = -1 х= π/9 - π/3 = π/9 - 3π/9 = -2π/9 ∉ [-5π/2; -2π],
при n = -2 х= π/9 - 2π/3 = π/9 - 6π/9 = -5π/9 ∉ [-5π/2; -2π],
при n = -3 х= π/9 - 3π/3 = π/9 - π = -8π/9 ∉ [-5π/2; -2π],
при n = -4 х= π/9 - 4π/3 =π/9 - 12π/9 = -11π/9 ∉ [-5π/2; -2π],
при n = -5 х= π/9 - 5π/3 = π/9 - 15π/9 = -14π/9 ∉ [-5π/2; -2π],
при n = -6 х= π/9 - 6π/3 = π/9 - 2π = -17π/9 ∉ [-5π/2; -2π],
при n = -7 х= π/9 - 7π/3 = π/9 - 21π/9 = -20π/9 ∈ [-5π/2; -2π],
при n = -8 х= π/9 - 8π/3 = π/9 - 24π/9 =  -23π/9 ∉ [-5π/2; -2π],

ответ: -20π/9.