Question

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

Answer 1

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.