(3ctg3x+√3)(tg4x+1)=0 произведение равно 0, когда один из множителей равен 0 3ctg3x+√3=0 tg4x+1=0 ctg3x=-√3/3 tg4x=-1 3x=π/6+πn 4x=3π/4+πn x=π/18+πn/3 x=3π/16+πn
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;
произведение равно 0, когда один из множителей равен 0
3ctg3x+√3=0 tg4x+1=0
ctg3x=-√3/3 tg4x=-1
3x=π/6+πn 4x=3π/4+πn
x=π/18+πn/3 x=3π/16+πn
ctg(x/2+π/4)-√3=0
ctg (x/2) *ctg(π/4) -1
= √3
ctg (x/2) - 1
1=√3 - решений не имеет
tg x =1 x∈[0;2π]
x=π/4 +πn
x1=π/4
x2=5π/4
(3ctg3x+√3) = 0
ctg3x = -
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;