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) = 0ctg3x = -/33x = π/3 + πn, n ∈ Zx = π/6 + π/3n, n ∈ Z(tg4x+1)=0tg4x = -1 4x = π/4 + πn, n ∈ Z;x = π/16 + π/4n, n ∈ Z2. ctg(x/2+пи/4)-√3=0ctg(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] тоx = π/4 + πn, n = 0,1;