Рeшите: ctg 2x + 4 = 0 2sin 2x + 1 = 0 tg 3x - 1 = 0 √2 cosx - 1 = 0

ctg 2x + 4 = 0;

ctg 2x = - 4;

2 х = arcctg (-4) + πk;

2 х = п - arcctg 4 + πk;

2 х = - arcctg 4 + п + πk;

2 х = - arcctg 4 + п (k + 1);

х = - 1/2 arcctg 4 + п/2 (k + 1), k ∈ Z;

2sin 2x + 1 = 0;

2sin 2x = - 1;

sin 2x = - 1/2;

2x = (-1) ^k arcsin (-1/2) + 2 пk;

2x = - (-1) ^k arcsin (1/2) + 2 пk;

2 х = - (-1) ^k п/6 + 2 Пk;

х = - (-1) ^k п/12 + Пk, k ∈ Z;

tg 3x - 1 = 0;

tg 3x = 1;

3x = arctg 1 + пk;

3x = п/4 + пk;

x = п/12 + пk/3, k ∈ Z;

√2 cos x - 1 = 0;

√2 cos x = 1;

cos x = 1/√2;

cos x = √2/2;

x = ± arccos √2/2 + 2 пk;

x = ± п/4 + 2 пk, k ∈ Z;