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.