Bonjour,
a) e^(ix) = cos(x) + i(sin(x)
Forme algébrique :
Z = (√3/2 - i/2)e^(ix)
= (√3/2 - i/2)(cos(x) + i(sin(x))
= √3/2 * cos(x) + 1/2 * sin(x) + [√3/2 * sin(x) - 1/2 * cos(x)]i
Forme exponentielle :
cos(-π/6) = √3/2 et sin(-π/6) = -1/2
⇒ (√3/2 - i/2) = e^(iπ/6)
⇒ Z = e^(iπ/6) x e^(ix) = e^i(x + π/6)
b) (E) : √3cos(x) + sin(x) = √2
⇔ √3/2 * cos(x) + 1/2 * sin(x) = √2/2
Z = R + Ii avec R partie réelle et I partie imaginaire de Z
Or : R = √3/2 * cos(x) + 1/2 * sin(x)
(E) ⇔ R = √2/2
Or : Z = e^i(x + π/6)
⇒ R = cos(x + π/6)
Donc : (E) ⇔ cos(x + π/6) = √2/2
⇔ cos(x + π/6) = cos(π/4)
⇒ x + π/6 = π/4 + k2π
ou x + π/6 = -π/4 + k2π
⇔ x = 3π/12 - 2π/12 + k2π
ou x = -3π/12 - 2π/12 + k2π
⇔ x = π/12 + k2π ou x = -5π/12 + k2π
A restreindre sur }-π; π]...
