Question

resolva, no intervalo 0<x<2pi, a equação 2senx(ao quadrado) - cosx - 1 =0

Answer 1

2sen²x - cosx -1 = 0
2(1-cos²x) - cosx - 1 = 0
2 - 2cos²x - cosx - 1 = 0
-2co²x - cosx + 1 = 0
2cos²x + cosx -1 = 0
Δ = 1 -4.2(-1)
Δ = 9
cosx = (-1 - 3)/4 => cos x = -1 => x = π rad ou
cosx = (-1 + 3)/4 => cosx = 1/2 => x = π/3 rad