Решите уравнение 4cosx ctgx + 4ctgx+sinx=0

4cosx ctgx + 4ctgx+sinx=0

4cosx·сosx/sinx + 4·сosx/sinx+sinx=0

4cos²x/sinx + 4·сosx/sinx+sin²x/sinx=0

sinx ≠ 0

4cos²x + 4·сosx+sin²x=0

3cos²x + 4·сosx+1=0

cosx = y

3y² + 4y+1=0

D = 16 - 12 = 4

x₁ = (-4 - 2) : 6 = - 1

x₂ = (-4 + 2) : 6 = - 1/3

cosx₁ = - 1

x₁ = π + 2πn

cosx₂ = - 1/3

x₂ = ±arccos (-1/3) + πn

4cos²x/sinx + 4cosx/sinx + sinx = 0

приводим к общему знаменателю

(4 сos²x + 4cosx + sin²x) / sinx = 0

sinx ≠ 0

(4 сos²x + 4cosx + sin²x) = 0

sin²x = 1-cos²x

4cos²x + 4cosx + 1 - cos²x

3cos²x + 4cosx + 1 = 0

cosx = a

3a² + 4a + 1 = 0

-4±√ (16 - 12) / 6

a1 = - 4 + 2 / 6 = - 1/3

a2 = - 4-2 / 6 = - 1

1) cosx = - 1/3

x = ± (π-arccos1/3) + 2πn

2) cosx = - 1

x = π + 2πn