Question

2cos^2 x+3sin2x-8sin^2x=0

Answer 1

2сos²x + 3sin2x - 8sin²x = 0
8sin²x - 3sin2x - 2cos²x = 0
8sin²x - 6sinxcosx - 2cos²x = 0      |:2cos²x, cosx ≠ 0
4tg²x - 3tgx - 1 = 0
Пусть t = tgx.
4t² - 3t - 1 = 0
D = 9 + 4·4 = 25 = 5²
t₁ = (3 + 5)/8 = 8/8 = 1
t₂ = (3 - 5)/8 = -2/8 = -1/4
Обратная замена:
tgx = 1
x = π/4 + πn, n ∈ Z
tgx = -1/4
x = arctg(-1/4) +  πk, k ∈ Z
Ответ: x = π/4 + πn, n ∈ Z; arctg(-1/4) +  πk, k ∈ Z.