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