Question

5-cosx=(-корень из 6 sinx)^2

Answer 1

5 - Cosx=(- √6 Sinx)^2
5 - Cosx = 6Sin²x
6Sin²x +Cosx -5= 0
6(1 - Cos²x) + Cosx -5 = 0
6 -6Cos²x +Cosx -5 = 0
6Cos²x -Cosx -1 = 0
Cosx = t
6t² -t -1 = 0
D = 1 -4*6*(-1) = 25
t₁ = (1 +5)/12 = 1/2;                 t₂ = (1 -5)/12 = -1/3
Сosx = 1/2                               Cosx = -1/3
x = +-π/3 +2πk , k ∈Z              x = +-arcCos(-1/3) + 2πn , n ∈ Z