Bonjour Abidol1
[tex]\cos^2(x+\dfrac{2\pi}{3})=\sin^2(x+\dfrac{2\pi}{3})\\\\\cos^2(x+\dfrac{2\pi}{3})=1-\cos^2(x+\dfrac{2\pi}{3})\\\\2\cos^2(x+\dfrac{2\pi}{3})=1\\\\\cos^2(x+\dfrac{2\pi}{3})=\dfrac{1}{2}\\\\\cos(x+\dfrac{2\pi}{3})=\pm\sqrt{\dfrac{1}{2}}=\pm\dfrac{\sqrt{2}}{2}[/tex]
D'une part :
[tex]\cos(x+\dfrac{2\pi}{3})=\dfrac{\sqrt{2}}{2}\\\\\cos(x+\dfrac{2\pi}{3})=\cos(\dfrac{\pi}{4})\\\\x+\dfrac{2\pi}{3}=\dfrac{\pi}{4}+2k\pi\ \ ou\ \ x+\dfrac{2\pi}{3}=-\dfrac{\pi}{4}+2k\pi\\\\x=\dfrac{\pi}{4}-\dfrac{2\pi}{3}+2k\pi\ \ ou\ \ x=-\dfrac{\pi}{4}-\dfrac{2\pi}{3}+2k\pi\\\\x=\dfrac{3\pi}{12}-\dfrac{8\pi}{12}+2k\pi\ \ ou\ \ x=-\dfrac{3\pi}{12}-\dfrac{8\pi}{12}+2k\pi\\\\\boxed{x=-\dfrac{5\pi}{12}+2k\pi\ \ ou\ \ x=-\dfrac{11\pi}{12}+2k\pi\ \ (k\ \in\ \mathbb{Z})}[/tex]
D'autre part,
[tex]\cos(x+\dfrac{2\pi}{3})=-\dfrac{\sqrt{2}}{2}\\\\\cos(x+\dfrac{2\pi}{3})=\cos(\dfrac{3\pi}{4})\\\\x+\dfrac{2\pi}{3}=\dfrac{3\pi}{4}+2k\pi\ \ ou\ \ x+\dfrac{2\pi}{3}=-\dfrac{3\pi}{4}+2k\pi\\\\x=\dfrac{3\pi}{4}-\dfrac{2\pi}{3}+2k\pi\ \ ou\ \ x=-\dfrac{3\pi}{4}-\dfrac{2\pi}{3}+2k\pi\\\\x=\dfrac{9\pi}{12}-\dfrac{8\pi}{12}+2k\pi\ \ ou\ \ x=-\dfrac{9\pi}{12}-\dfrac{8\pi}{12}+2k\pi\\\\x=\dfrac{\pi}{12}+2k\pi\ \ ou\ \ x=-\dfrac{17\pi}{12}+2k\pi[/tex]
[tex]x=\dfrac{\pi}{12}+2k\pi\ \ ou\ \ x=-\dfrac{17\pi}{12}+2\pi+2k\pi\\\\x=\dfrac{\pi}{12}+2k\pi\ \ ou\ \ x=-\dfrac{17\pi}{12}+\dfrac{24\pi}{12}+2k\pi\\\\\boxed{x=\dfrac{\pi}{12}+2k\pi\ \ ou\ \ x=\dfrac{7\pi}{12}+2k\pi\ \ (k\ \in\ \mathbb{Z})}[/tex]
La représentation graphique est en pièce jointe.
