solution en [-π , π] -2 sin²x + ( 1-[tex] \sqrt{3} [/tex]) sinx + [tex] \sqrt{3}[/tex]\2 [tex] \geq [/tex] 0

-2sin²(x)+(1-√3)sin(x)+√3/2 ≥ 0
sin²(x)-(1-√3)/2.sin(x)-√3/4 ≤ 0
(sin(x)-1/2)(sin(x)+√3/2) ≤ 0
-√3/2 ≤ sin(x) ≤ 1/2
-π/3 ≤ x ≤ π/6 ou 5π/6 ≤ x ≤ 4π/3