Bonjour,
a) On pose X = cos(x)
L'équation devient : 2X² + 9X + 4 = 0
Δ = 9² - 4x2x4 = 81 - 32 = 49 = 7²
donc 2 solutions :
X₁ = (-9 - 7)/4 = -4
X₂ = (-9 + 7)/4 = -1/2
On en déduit :
cos(x) = -4 pas de solution car ∀ x ∈ R, -1 ≤ cos(x) ≤ 1
cos(x) = -1/2
⇔ cos(x) = cos(2π/3)
⇒ x = 2π/3 + k2π ou x = -2π/3 + k2π
b) idem, on pose X = sin(x)
4X² - 2(1 + √3)X + √3 = 0
Δ = [-2(1 + √3)]² - 4x4x√3
⇔ Δ = 4(1 + 2√3 + 3) - 16√3
⇔ Δ = 4(1 - 2√3 + 3)
⇔ Δ = 4(1 - √3)² = [2(1 - √3)]²
X₁ = [2(1 + √3) - 2(1 - √3)]/8 = √3/2
X₂ = [2(1 + √3) + 2(1 - √3)]/8 = 1/2
sin(x) = √3/2
⇔ sin(x) = sin(π/3)
⇒ x = π/3 + k2π et x = 2π/3 + k2π
sin(x) = 1/2
⇔ sin(x) = sin(π/6)
⇒ x = π/6 + k2π et x = 5π/6 + k2π
