Question

5-4 cos2 x = 5 sin2 x

Answer 1

5 - (4 - 8sin²x) = 10sinxcosx
1 + 8sin²x = 10sinxcosx
9sin²x + cos²x - 10sinxcosx = 0
9tg²x - 10tgx + 1 = 0
Пусть t = tgx.
9t² - 10t + 1 = 0
D = 100 - 36 = 64 = 8²
t1 = (10 + 8)/18 = 1
t2 = (10 - 8)/18 = 2/18 = 1/9.
Обратная замена:
tgx = 1
x = π/4 + 2πn, n ∈ Z.
tgx = 1/9
x = arctg(1/9) + πn, n ∈ Z.