Решение
7sin^2x=8sinxcosx-cos^2x
7sin^2x - 8sinxcosx + cos^2x = 0 делим на cos²x ≠ 0
7tg²x - 8tgx + 1 = 0
tgx = t
7t² - 8t + 1 = 0
D = 64 - 4*7*1 = 36
t = (8 - 6)/14 = - 1/7
t = (8 + 6)/14 = 1
1) tgx = - 1/7
x₁ = - arctg(1/7) + πk, k ∈ Z
2) tgx = 1
x₂ = π/4 + πn, n ∈Z
Ответ: x₁ = - arctg(1/7) + πk, k ∈ Z ; x₂ = π/4 + πn, n ∈Z