[tex] :cos^2 (x+2π/3)=sin²(x+2π/3) [/tex] bonjour un peux largue, sa remonte a trop longtemps  lol J'ai donc a résoudre l'aquation ci-desus Et construire les ex

Bonsoir,

[tex]\cos^2(x+\dfrac{\pi}{3})=\sin^2(x+\dfrac{\pi}{3})\\\\\cos^2(x+\dfrac{\pi}{3})=1-\cos^2(x+\dfrac{\pi}{3})\\\\2\cos^2(x+\dfrac{\pi}{3})=1\\\\\cos^2(x+\dfrac{\pi}{3})=\dfrac{1}{2}\\\\\cos(x+\dfrac{\pi}{3})=\pm\sqrt{\dfrac{1}{2}}\\\\\cos(x+\dfrac{\pi}{3})=\pm\dfrac{\sqrt{2}}{2}[/tex]

[tex]a)\ \cos(x+\dfrac{\pi}{3})=\dfrac{\sqrt{2}}{2}\\\\\cos(x+\dfrac{\pi}{3})=\cos(\dfrac{\pi}{4})\\\\x+\dfrac{\pi}{3}=\dfrac{\pi}{4}+2k\pi\ \ ou\ \ x+\dfrac{\pi}{3}=-\dfrac{\pi}{4}+2k\pi\\\\x=\dfrac{\pi}{4}-\dfrac{\pi}{3}+2k\pi\ \ ou\ \ x=-\dfrac{\pi}{4}-\dfrac{\pi}{3}+2k\pi\\\\x=\dfrac{3\pi}{12}-\dfrac{4\pi}{12}+2k\pi\ \ ou\ \ x=-\dfrac{3\pi}{12}-\dfrac{4\pi}{12}+2k\pi\\\\x=-\dfrac{\pi}{12}+2k\pi\ \ ou\ \ x=-\dfrac{7\pi}{12}+2k\pi[/tex]

[tex]b)\ \cos(x+\dfrac{\pi}{3})=-\dfrac{\sqrt{2}}{2}\\\\\cos(x+\dfrac{\pi}{3})=\cos(\dfrac{3\pi}{4})\\\\x+\dfrac{\pi}{3}=\dfrac{3\pi}{4}+2k\pi\ \ ou\ \ x+\dfrac{\pi}{3}=-\dfrac{3\pi}{4}+2k\pi\\\\x=\dfrac{3\pi}{4}-\dfrac{\pi}{3}+2k\pi\ \ ou\ \ x=-\dfrac{3\pi}{4}-\dfrac{\pi}{3}+2k\pi\\\\x=\dfrac{9\pi}{12}-\dfrac{4\pi}{12}+2k\pi\ \ ou\ \ x=-\dfrac{9\pi}{12}-\dfrac{4\pi}{12}+2k\pi\\\\x=\dfrac{5\pi}{12}+2k\pi\ \ ou\ \ x=-\dfrac{13\pi}{12}+2k\pi[/tex]

Donc   [tex]x=-\dfrac{\pi}{12}+k\dfrac{\pi}{2}[/tex]

Les extrémités des arcs solutions sont les points A, B, C et D (pièce jointe)