Помогите решить уравнение 1-5sin2x-cos2x=12cos^x

1) 5Sin²x - Cos²x = 4 + 4Sinx*Cosx
5Sin²x - Cos²x = 4Sin²x + 4Cos²x + 4Sinx*Cosx
5Sin²x - Cos²x - 4Sin²x - 4Cos²x - 4Sinx*Cosx = 0
Sin²x - 4Sinx*Cosx - 5Cos²x = 0 / Cos²x
tg²x - 4tgx - 5 = 0

1) Замена: tgx = t
t² - 4t - 5 = 0
t₁ = -1
t₂ = 5

2) tgx = -1
x = -π/4 + π*n, n ∈ Ζ

tgx = 5
x = arctg5 + π*n, n ∈ Ζ

2) 3 + 2Sin3x*Sinx = 3Cos2x
3Sin²x + 3Cos²x + 2Sinx(3Sinx - 4Sin³x) - 3Cos²x + 3Sin²x = 0
6Sin²x + 6Sin²x - 8Sin⁴x = 0
12Sin²x - 8Sin⁴x = 0

1) Замена: Sin²x = t
12t - 8t² = 0
t₁ = 0
t₂ = 3/2

2) Sin²x = 0
x = π*n, n ∈ Z

Sin²x = 3/2
x = ± √(3/2) + 2π*n, n ∈ Z