Помогите!!!
1) соs x * tg 3x = 0
2) 8 + sin 2x= 14 cos^2 x

1) cosx•tg3x = 0cosx = 0x = π/2 + πn, n ∈ Ztg3x = 03x = πn, n ∈ Zx = πn/3, n ∈ Z2) 8 + sin2x = 14cos²x8sin²x + 8cos²x + 2sinxcosx - 14cos²x = 08sin²x + 2sinxcosx - 6cos²x = 04tg²x + tgx - 3 = 0Пусть t = tgx.4t² + t - 3 = 0D = 1 + 3•4•4 = 49 = 7²t1 = (-1 + 7)/8 = 6/8 = 3/4t2 = (-1 - 7)/8 = -1Обратная замена:tgx = 3/4x = arctg(3/4) + πn, n ∈ Ztgx = -1x = -π/4 + πn, n ∈ Z.