1) cos (1-x) = 1/2 2) cosx/4=4/5 3) tg2x=√3 4) tg (3x+) = √3/2

1)

cos (1 - x) = 1/2; cos (x - 1) = 1/2; x - 1 = ±π/3 + 2πk, k ∈ Z; x = 1 ± π/3 + 2πk, k ∈ Z.

2)

cos (x/4) = 4/5; x/4 = ±arccos (4/5) + 2πk, k ∈ Z; x = ±4arccos (4/5) + 8πk, k ∈ Z.

3)

tg2x = √3; 2x = π/3 + πk, k ∈ Z; x = π/6 + πk/2, k ∈ Z.

4)

tg (3x + 1) = √3/2; 3x + 1 = arctg (√3/2) + πk, k ∈ Z; 3x = arctg (√3/2) - 1 + πk, k ∈ Z; x = arctg (√3/2) / 3 - 1/3 + πk/3, k ∈ Z.

Ответ:

1) 1 ± π/3 + 2πk, k ∈ Z; 2) ±4arccos (4/5) + 8πk, k ∈ Z; 3) π/6 + πk/2, k ∈ Z; 4) arctg (√3/2) / 3 - 1/3 + πk/3, k ∈ Z.