Question

Решите уравнение $5sin^{2}x+4cosx=4$

Answer 1

5(1 - Cos²x) +4Cosx -4 = 0

5 - 5Cos²x +4Cosx -4 = 0

5Cos²x -4Cosx -1 = 0

Cosx = t

5t² -4t -1 = 0

t =(2 +-√9)/5

t₁ =(2 +3)/5 = 1                    t₂= (2 -3)/5

Cosx = 1                               Cosx = -1/5

x = 2πk, k∈Z                         x = +-arcCos(-0,2) +2πn , n∈Z