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Kingkylie21
@Kingkylie21
May 2019
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Bonjour , j’ai un dm de math à faire je comprend pas du tout , j’aimerais de l’aide , merci d’avance c’est pour dans 3 jours
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scoladan
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Bonjour,
Ex 2)
sin(2x - π/2) = sin(x)
⇒ 2x - π/2 = x + k2π avec k ∈ Z
ou 2x - π/2 = (π - x) + k'2π avec k' ∈ Z
⇔ x = π/2 + k2π
ou 3x = 3π/2 + k'2π ⇔ x = π/2 + k'2π/3
Sur I = [-π;π] :
(k = -1 ⇒ x = -3π/2 ∉ I)
k = 0 ⇒ x = π/2
(k = 1 ⇒ x = 5π/2 ∉ I)
(k' = -3 ⇒ x = π/2 - 2π = -3π/2 ∉ I)
k' = -2 ⇒ x = π/2 -4π/3 = 3π/6 - 8π/6 = -5π/6
k' = -1 ⇒ x = π/2 - 2π/3 = 3π/6 - 4π/6 = -π/6
k' = 0 ⇒ x = π/2
(k' = 1 ⇒ x = π/2 + 2π/3 = 7π/6 ∉ I)
Donc les solutions sont : -5π/6, -π/6 et π/2
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kingkylie21
Merci infiniment !
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Bonjour, j'ai un dm de math que je doit rendre pour demain qu'on ma donner seulement hier et j'aimerais que l'on m'aide .
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Verified answer
Bonjour,Ex 2)
sin(2x - π/2) = sin(x)
⇒ 2x - π/2 = x + k2π avec k ∈ Z
ou 2x - π/2 = (π - x) + k'2π avec k' ∈ Z
⇔ x = π/2 + k2π
ou 3x = 3π/2 + k'2π ⇔ x = π/2 + k'2π/3
Sur I = [-π;π] :
(k = -1 ⇒ x = -3π/2 ∉ I)
k = 0 ⇒ x = π/2
(k = 1 ⇒ x = 5π/2 ∉ I)
(k' = -3 ⇒ x = π/2 - 2π = -3π/2 ∉ I)
k' = -2 ⇒ x = π/2 -4π/3 = 3π/6 - 8π/6 = -5π/6
k' = -1 ⇒ x = π/2 - 2π/3 = 3π/6 - 4π/6 = -π/6
k' = 0 ⇒ x = π/2
(k' = 1 ⇒ x = π/2 + 2π/3 = 7π/6 ∉ I)
Donc les solutions sont : -5π/6, -π/6 et π/2