Помогите решить:

1) cos (x+п/4) = 1

2) sin (3 п/2-x) = - 1

3) sin (-x) = - 0,5

4) tgx/2=корень из 3

5) cos (2x-п/3) = корень из3/2

V

1) x+π/4 = 2π*k, k∈ Z.

x = - π/4 x+2π*k, k∈ Z.

2) sin (3π/2 - x) = - 1;

- cosx = - 1;

cosx = 1;

x=2π*k, k ∈ Z.

3) sin (-x) = - 1/2;

- sinx = - 1/2

sinx = 1/2;

x = (-1) ^ (k) * π/6+π*k, k ∈Z.

4) tq (x/2) = √3; [ (tqx) / 2 = √3 ];

x/2 = π/3 + π*k, k ∈ Z.

x=2π/3 + 2π*k, k ∈ Z.

5) cos (2x-π/3) = (√3) / 2;

[2x - π/3 = - π/6 + 2π*k, k ∈ Z; 2x - π/3 = π/6 + 2π*k, k ∈ Z.

[2x = π/3 - π/6 + 2π*k, k ∈ Z; 2x = π/3 + π/6 + 2π*k, k ∈ Z

[2x = π/6 + 2π*k, k ∈ Z; 2x = π/2 + 2π*k, k ∈ Z.

x = π/12 + π*k, k ∈ Z;

x = π/4 + π*k, k ∈ Z.