4 COS ^ 2x + senx - 1 = 0 (Ec)?
4 COS ^ 2x + senx - 1 = 0 (Ec).
2Mel132
En resumen
Cos²x = 1 - sen²x 4(1 - sen²x) + senx - 1 = 0 4 - 4sen²x + senx - 1 = 0 4sen²x - senx - 3 = 0 4senx 3 senx - 1 (4senx + 3)(senx - 1) = 0 senx = - 3 / 4 ∨ senx = 1 SALUDOS ISABELA.
Mejor respuesta
CaIp
Cos²x = 1 - sen²x
4(1 - sen²x) + senx - 1 = 0
4 - 4sen²x + senx - 1 = 0
4sen²x - senx - 3 = 0
4senx 3
senx - 1
(4senx + 3)(senx - 1) = 0
senx = - 3 / 4 ∨ senx = 1
SALUDOS ISABELA.