Comment resoudre: Cos X = Sin X , et Sin (Pi/3-X) = Sin (2X+(Pi/4)) ?
Merci d'avance à ceux qui regarderons.
cosx = cox(pi/2 - x) => x = + ou - (pi/2 - x) + k2pi
=> x = pi/2 - x+ k2pi=>2x = pi/2 + k2pi => x = pi/4 + k2pi
=> x = -pi/2 + x + k2pi => 0x = -pi/2+ k2pi impossible
Sin ) = Sin (Pi/3-X)
(2X+(Pi/4) = (Pi/3-X) + k2pi => 2x + pi/4 = pi/3 - x + k2pi => 3x = pi/3 - pi/4 + k2pi
=> 3x = pi/12 + k2pi => x = pi/36 + k2pi/3 (3solutions)
ou bien
2x + pi/4 = pi - pi/3 + x + k2pi => x = 2pi/3 - pi/4 + k2pi => x = 5pi/12 + k2pi
Copyright © 2024 ELIBRARY.TIPS - All rights reserved.
Lista de comentários
Verified answer
cosx = cox(pi/2 - x) => x = + ou - (pi/2 - x) + k2pi
=> x = pi/2 - x+ k2pi=>2x = pi/2 + k2pi => x = pi/4 + k2pi
=> x = -pi/2 + x + k2pi => 0x = -pi/2+ k2pi impossible
Sin ) = Sin (Pi/3-X)
(2X+(Pi/4) = (Pi/3-X) + k2pi => 2x + pi/4 = pi/3 - x + k2pi => 3x = pi/3 - pi/4 + k2pi
=> 3x = pi/12 + k2pi => x = pi/36 + k2pi/3 (3solutions)
ou bien
2x + pi/4 = pi - pi/3 + x + k2pi => x = 2pi/3 - pi/4 + k2pi => x = 5pi/12 + k2pi