решите уравнения
(3ctg3x+√3)(tg4x+1)=0
ctg(x/2+пи/4)-√3=0
tgx=1 x ∈[0;2пи]

1. (3ctg3x+√3)(tg4x+1)=0
(3ctg3x+√3) = 0
ctg3x = -/3
3x = π/3 + πn, n ∈ Z
x = π/6 + π/3n, n ∈ Z
(tg4x+1)=0
tg4x = -1 
4x = π/4 + πn, n ∈ Z;
x = π/16 + π/4n, n ∈ Z

2. ctg(x/2+пи/4)-√3=0
ctg(x/2+пи/4) = 
x/2 + π/4 = π/6 + πn, n ∈ Z;
x = π/3 - π/2 + 2πn,n ∈ Z'
x = -π/6 + 2πn, n ∈ Z;

3.tgx=1 x ∈[0;2пи]
x = π/4 + πn,n ∈ Z;
Поскольку x ∈[0;2пи] то
x = π/4 + πn, n = 0,1;