Bonjour,
Utilisons une formule de duplication :
[tex]\cos(2x)=\cos^2x-\sin^2x[/tex]
[tex]\cos(2x)=\cos^2x-(1-\cos^2x)\\\\\cos(2x)=\left(\dfrac{\sqrt{6}+\sqrt{2}}{4}\right)^2-\left(1-\left(\dfrac{\sqrt{6}+\sqrt{2}}{4}\right)^2\right)\\\\\cos(2x)=\dfrac{\left(\sqrt{6}+\sqrt{2}\right)^2}{16}-\left(1-\dfrac{\left(\sqrt{6}+\sqrt{2}\right)^2}{16}\right)\\\\\cos(2x)=\dfrac{6+2\sqrt{12}+2}{16}-\left(1-\dfrac{6+2\sqrt{12}+2}{16}\right)\\\\\cos(2x)=\dfrac{8+4\sqrt{3}}{16}-\left(1-\dfrac{8+4\sqrt{3}}{16}\right)\\\\\cos(2x)=\dfrac{2+\sqrt{3}}{4}-\left(1-\dfrac{2+\sqrt{3}}{4}\right)\\\\\cos(2x)=\dfrac{2+\sqrt{3}}{4}-\dfrac{4-2-\sqrt{3}}{4}\\\\\cos(2x)=\dfrac{2+\sqrt{3}}{4}-\dfrac{2-\sqrt{3}}{3}\\\\\cos(2x)=\dfrac{2\sqrt{3}}{4}\\\\\boxed{\cos(2x)=\dfrac{\sqrt{3}}{2}}[/tex]
[tex]\cos(2x)=\dfrac{\sqrt{3}}{2}\\\\2x=\dfrac{\pi}{6}+2k\pi\text{ ou }2\pi-2x=\dfrac{\pi}{6}\\\\\\x=\dfrac{\pi}{12}+k\pi\text{ ou }x=\dfrac{11\pi}{12}+k\pi\\\\\text{ or }0\ \textless \ x\ \textless \ \dfrac{\pi}{2}\\\\\text{donc}\\\\\boxed{x=\dfrac{\pi}{12}+k\pi}[/tex]
