Помогите
5sin^2X-2*COSX+COS^2X=4

Решение
5sin²x - 2cosx + cos²x = 4
5*(1 - cos²x) - 2cosx + cos²x - 4 = 0
5 - 5 cos²x - 2cosx + cos²x - 4 = 0
- 4cos²x - 2cosx + 1 = 0
4cos²x + 2cosx - 1 = 0
cosx = t
4t² + 2t - 1 = 0
D = 4 + 4*4*1 = 20
t₁ = (- 2 - 2√5)/8
t₁ = (- 1 -  √5)/4
t₂ = (- 1 + √5)/4
1) cosx = (- 1 -  √5)/4
x = (+ -)arccos((- 1 -  √5)/4 + 2πk, k∈Z
cosx = (- 1 +  √5)/4
x =  (+ -)arccos((- 1 +  √5)/4 + 2πn, n∈Z