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

(4сos²x + 4cosx + sin²x)/sinx = 0

sinx ≠ 0 

(4сos²x + 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