Помогите!

1) соs x * tg 3x = 0

2) 8 + sin 2x = 14 cos^2 x

1) cosx•tg3x = 0

cosx = 0

x = π/2 + πn, n ∈ Z

tg3x = 0

3x = πn, n ∈ Z

x = πn/3, n ∈ Z

2) 8 + sin2x = 14cos²x

8sin²x + 8cos²x + 2sinxcosx - 14cos²x = 0

8sin²x + 2sinxcosx - 6cos²x = 0

4tg²x + tgx - 3 = 0

Пусть t = tgx.

4t² + t - 3 = 0

D = 1 + 3•4•4 = 49 = 7²

t1 = (-1 + 7) / 8 = 6/8 = 3/4

t2 = (-1 - 7) / 8 = - 1

Обратная замена:

tgx = 3/4

x = arctg (3/4) + πn, n ∈ Z

tgx = - 1

x = - π/4 + πn, n ∈ Z.