Mathématiques
martin15
15

svp j'ai besoin de résoudre les équation suivante a) cos [latex] (3x - \frac{\pi}{4} ) = \frac{ \sqrt{3} }{2} [/latex] b) cos (x) = [latex]- \frac{1}{2} [/latex] c) cos (2x [latex]- \pi ) = sin ( \frac{ \pi }{4} )[/latex] d)sin (3x -[latex] \pi ) = cos ( \frac{2 \pi }{3} )[/latex] SVP c'est URGENT !!!

+1
(1) Réponses
elidrissianissa

Bonjour  Souleym [latex]a)\ \cos(3x-\dfrac{\pi}{4})=\dfrac{\sqrt{3}}{2}\\\\\cos(3x-\dfrac{\pi}{4})=\cos(\dfrac{\pi}{6})\\\\3x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+2k\pi\ \ ou\ \ 3x-\dfrac{\pi}{4}=-\dfrac{\pi}{6}+2k\pi\\\\3x=\dfrac{\pi}{6}+\dfrac{\pi}{4}+2k\pi\ \ ou\ \ 3x=-\dfrac{\pi}{6}+\dfrac{\pi}{4}+2k\pi\\\\3x=\dfrac{2\pi}{12}+\dfrac{3\pi}{12}+2k\pi\ \ ou\ \ 3x=-\dfrac{2\pi}{12}+\dfrac{3\pi}{12}+2k\pi\\\\3x=\dfrac{5\pi}{12}+2k\pi\ \ ou\ \ 3x=\dfrac{\pi}{12}+2k\pi[/latex] [latex]\boxed{x=\dfrac{5\pi}{36}+\dfrac{2k\pi}{3}\ \ ou\ \ x=\dfrac{\pi}{36}+\dfrac{2k\pi}{3}\ \ (k\in\mathbb{Z})}[/latex] [latex]b)\ \cos(x)=-\dfrac{1}{2}\\\\\cos(x)=\cos(\dfrac{2\pi}{3})\\\\\boxed{x=\dfrac{2\pi}{3}+2k\pi\ \ ou\ \ x=-\dfrac{2\pi}{3}+2k\pi\ \ (k\in\mathbb{Z})}[/latex] [latex]c)\ \cos(2x-\pi)=\sin(\dfrac{\pi}{4})\\\\\cos(2x-\pi)=\cos(\dfrac{\pi}{2}-\dfrac{\pi}{4})\\\\\cos(2x-\pi)=\cos(\dfrac{\pi}{4})\\\\2x-\pi=\dfrac{\pi}{4}+2k\pi\ \ ou\ \ 2x-\pi=-\dfrac{\pi}{4}+2k\pi\\\\2x=\dfrac{\pi}{4}+\pi+2k\pi\ \ ou\ \ 2x=-\dfrac{\pi}{4}+\pi+2k\pi\\\\2x=\dfrac{5\pi}{4}+2k\pi\ \ ou\ \ 2x=\dfrac{3\pi}{4}+2k\pi\\\\\boxed{x=\dfrac{5\pi}{8}+k\pi\ \ ou\ \ x=\dfrac{3\pi}{8}+k\pi\ \ (k\in\mathbb{Z})}[/latex] [latex]d)\ \sin(3x-\pi)=\cos(\dfrac{2\pi}{3})\\\\\cos[\dfrac{\pi}{2}-(3x-\pi)]=\cos(\dfrac{2\pi}{3})\\\\\cos(\dfrac{\pi}{2}-3x+\pi)=\cos(\dfrac{2\pi}{3}) \\\\\cos(\dfrac{3\pi}{2}-3x)=\cos(\dfrac{2\pi}{3})\\\\\cos(3x-\dfrac{3\pi}{2})=\cos(\dfrac{2\pi}{3})[/latex] [latex]3x-\dfrac{3\pi}{2}=\dfrac{2\pi}{3}+2k\pi\ \ ou\ \ 3x-\dfrac{3\pi}{2}=-\dfrac{2\pi}{3}+2k\pi\\\\3x=\dfrac{2\pi}{3}+\dfrac{3\pi}{2}+2k\pi\ \ ou\ \ 3x=-\dfrac{2\pi}{3}+\dfrac{3\pi}{2}+2k\pi\\\\3x=\dfrac{4\pi}{6}+\dfrac{9\pi}{6}+2k\pi\ \ ou\ \ 3x=-\dfrac{4\pi}{6}+\dfrac{9\pi}{6}+2k\pi\\\\3x=\dfrac{13\pi}{6}+2k\pi\ \ ou\ \ 3x=\dfrac{5\pi}{6}+2k\pi\\\\\boxed{x=\dfrac{13\pi}{18}+\dfrac{2k\pi}{3}\ \ ou\ \ x=\dfrac{5\pi}{18}+\dfrac{2k\pi}{3}\ \ (k\in\mathbb{Z})}[/latex]

Ajouter une réponse