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Ledernierdelaclasse
@Ledernierdelaclasse
May 2019
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Bonsoir à tous et à toutes, j'ai un problème pour cette équation sur la Trigonométrie, quelqu'un pourrait m'expliquer comment la résoudre ? J'ai vraiment
envie de comprendre :) Merci pour votre aide les amis ! :)
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AhYan
Bonsoir,
cos(x/3+pi/4) = 1/2
Or on sait que : cos(pi/3) = 1/2 et cos(-pi/3) = 1/2
Donc on a :
1- cos(x/3+pi/4) = cos(pi/3) donc (x/3+pi/4) = pi/3+2k*pi
2- cos(x/3+pi/4) = cos(-pi/3) donc (x/3+pi/4) = -pi/3+2k*pi
On a alors :
1- x/3+pi/4 = pi/3+2k*pi
==> x/3 = pi/3-pi/4+2k*pi
==> x/3 = (4pi - 3pi)/12+2k*pi
==> x/3 = pi/12 + 2k*pi
==> x = 3(pi/12+2k*pi)
==> x = 3pi/12 + 6k*pi
==> x = pi/4 + 6k*pi
2- (x/3+pi/4) = -pi/3+2k*pi
==> x/3 = -pi/3-pi/4+2k*pi
==> x/3 = (-4pi-3pi)/12 + 2k*pi
==> x/3 = -7pi/12 + 2k*pi
==> x = 3(-7pi/12 + 2k*pi)
==> x = -21pi/12 + 6k*pi
==> x = -7pi/4 + 6k*pi
1 votes
Thanks 1
AhYan
de rien ^^
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Lista de comentários
cos(x/3+pi/4) = 1/2
Or on sait que : cos(pi/3) = 1/2 et cos(-pi/3) = 1/2
Donc on a :
1- cos(x/3+pi/4) = cos(pi/3) donc (x/3+pi/4) = pi/3+2k*pi
2- cos(x/3+pi/4) = cos(-pi/3) donc (x/3+pi/4) = -pi/3+2k*pi
On a alors :
1- x/3+pi/4 = pi/3+2k*pi
==> x/3 = pi/3-pi/4+2k*pi
==> x/3 = (4pi - 3pi)/12+2k*pi
==> x/3 = pi/12 + 2k*pi
==> x = 3(pi/12+2k*pi)
==> x = 3pi/12 + 6k*pi
==> x = pi/4 + 6k*pi
2- (x/3+pi/4) = -pi/3+2k*pi
==> x/3 = -pi/3-pi/4+2k*pi
==> x/3 = (-4pi-3pi)/12 + 2k*pi
==> x/3 = -7pi/12 + 2k*pi
==> x = 3(-7pi/12 + 2k*pi)
==> x = -21pi/12 + 6k*pi
==> x = -7pi/4 + 6k*pi