Пожалуйста ,решите первые 5 номеров))) Я плохо поняла эту тему(((

1)tg² x = 3

tg x = √3                             или                                          tg x = -√3

x = arctg √3 + πn, n∈ Z                                                     x = arctg(-√3) + πk, k∈Z

x = π/3 + πn, n∈Z                                                               x = -π/3 + πk, k∈Z

Ответ: π/3 + πn, n∈Z; -π/3 + πk, k∈Z

2)ctg²x = 1/3

    ctg x = √3/3                                         или                                     ctg x = -√3/3

    x = arcctg √3/3 + πn,n∈Z                                                              x = arcctg(-√3/3) + πk, k∈Z

    x = π/3 + πn, n∈Z                                                                           x = 2π/3 + πk, k∈Z

 Ответ: π/3 + πn, n∈Z ; 2π/3 + πk, k∈Z.

3)cos²x = 3/4

    cos x = √3/2                                         или                                   cos x = -√3/2

    x = ±arccos √3/2 + 2πn, n∈Z                                                      x = ±arccos(-√3/2) + 2πk,k∈Z

   x = ±π/6 + 2πn, n∈ Z                                                                     x = ±5π/6 + 2πk,k∈Z

  Ответ: ±π/6 + 2πn, n∈ Z; ±5π/6 + 2πk,k∈Z

  4)sin²x = 1/2

     sin x = √2/2                                                 или                                    sin x = -√2/2

     x = (-1)^k * arcsin √2/2 + πn, n∈Z                                                      x = (-1)^k * arcsin(-√2/2)+πk

     x = (-1)^k * π/4 + πn, n∈Z                                                                    x = (-1)^k+1 * π/4 + πk,k∈Z

Ответ:(-1)^k * π/4 + πn, n∈Z; (-1)^k+1 * π/4 + πk,k∈Z

5)|tg x| = 4

   tg x = -4                                      или                                tg x = 4

   x = arctg (-4) + πn,n∈Z                                                   x = arctg 4 + πk, k∈Z

   x = -arctg 4 + πn, n∈Z

Ответ: -arctg 4 + πn, n∈Z; arctg 4 + πk, k∈Z