Question

3 tg(x-п/4)=√3
помогите пожалуйста решить

Answer 1

√3 tg(3x-π/4)+1≤0;tg(3x-π/4)≤-1/√3;-π/2+πn≤3x-π/4≤arctg(-1/√3)+πn, n∈Z;-π/2+πn ≤ 3x-π/4 ≤- π/6+πn, n∈Z;-π/2+π/4+πn ≤ 3x ≤ - π/6+π/4+πn, n∈Z;-π/6+π/12+(π/3)·n ≤ x ≤ - π/18+π/12+(π/3)·n, n∈Z;-π/12+(π/3)·n ≤ x ≤ π/36+(π/3)·n, n∈Z;