Exercice : Résoudre 1

Question

\(1. sin(3x+\frac{\pi}{2}) =-1 \quad x\in]0 ;2\pi]\)

Solution

\(sin(3x+\frac{\pi}{2}) =sin(\frac{3\pi}{2})\)

\(\begin{cases}3x+\frac{\pi}{2}=\frac{3\pi}{2}+2k\pi\\3x+\frac{\pi}{2}=\pi-\frac{3\pi}{2}+2k\pi\end{cases}\)

\(\iff \begin{cases}3x=\frac{3\pi}{2}-\frac{\pi}{2}+2k\pi\\3x+\frac{\pi}{2}=-\frac{\pi}{2}+2k\pi\end{cases}\)

\(\iff \begin{cases}3x=\frac{3\pi}{2}-\frac{\pi}{2}+2k\pi\\3x=-\frac{\pi}{2}-\frac{\pi}{2}+2k\pi\end{cases}\)

\(\iff \begin{cases}3x=\pi+2k\pi\\3x=-\pi+2k\pi\end{cases}\)

\(\iff \begin{cases}x=\frac{\pi}{3}+\frac{2k\pi}{3}\\x=-\frac{\pi}{3}+\frac{2k\pi}{3}\end{cases}\)

  • \(k=0\)

    \(x=\frac{\pi}{3} \in]-\pi ;\pi]\)

    \(x=\frac{-\pi}{3}\in]-\pi ;\pi]\)

  • \(k=1\)

    \(x=\frac{\pi}{3}+\frac{2\pi}{3}=\frac{3\pi}{3}=\pi\in]-\pi ;\pi]\)

    \(x=\frac{-\pi}{3}+\frac{2\pi}{3}=\frac{\pi}{3}\in]-\pi ;\pi]\)

  • \(k=2\)

    \(x=\frac{\pi}{3}+\frac{4\pi}{3}=\frac{5\pi}{3}\notin]-\pi ;\pi]\)

    Il est donc inutile de prendre des valeurs de \(k\) plus grande pour cette première solution.

    \(x=-frac{-\pi}{3}+\frac{4\pi}{3}=\frac{3\pi}{3}=\pi\in]-\pi ;\pi]\)

  • \(k=3\)

    \(x=\frac{-\pi}{3}+\frac{6\pi}{3}=\frac{5\pi}{3}\notin]-\pi ;\pi]\)

    Il est donc inutile de prendre des valeurs de \(k\) plus grande pour cette deuxième solution.

  • \(k=-1\)

    \(x=\frac{\pi}{3}+\frac{-2\pi}{3}=\frac{-\pi}{3}\in]-\pi ;\pi]\)

    \(x=\frac{-\pi}{3}+\frac{-2\pi}{3}=-\pi\notin]-\pi ;\pi]\)

    Il est donc inutile de prendre des valeurs de \(k\) plus petite pour cette deuxième solution.

  • \(k=-2\)

    \(x=\frac{\pi}{3}+\frac{-4\pi}{3}=\frac{-3\pi}{3}=-\pi\notin]-\pi ;\pi]\)

    Il est donc inutile de prendre des valeurs de \(k\) plus petite pour cette solution

Question

\(2.cos(3x)= \frac{-1}{2} \quad x\in]-\pi ;\pi]\)

Solution

\(cos(3x)= \frac{-1}{2}\)

\(\iff cos(3x)= cos(\pi-\frac{\pi}{3})\)

\(\iff cos(3x)= cos(\frac{3\pi}{3}-\frac{\pi}{3})\)

\(\iff cos(3x)= cos(\frac{2\pi}{3})\)

\(\begin{cases}3x=\frac{2\pi}{3}+2k\pi\\3x=-\frac{2\pi}{3}+2k\pi\end{cases}\)

\(\begin{cases}x=\frac{2\pi}{9}+\frac{2k\pi}{3}\\x=-\frac{2\pi}{9}+\frac{2k\pi}{3}\end{cases}\)

  • \(k=0\)

    \(x=\frac{2\pi}{9} \in]-\pi ;\pi]\)

    \(x=frac{-2\pi}{9}\in]-\pi ;\pi]\)

  • \(k=1\)

    \(x=\frac{2\pi}{9}+\frac{2\pi}{3}=\frac{2\pi}{9}+\frac{6\pi}{9}=\frac{8\pi}{9}\in]-\pi ;\pi]\)

    \(x=-frac{2\pi}{9}+\frac{2\pi}{3}=\frac{-2\pi}{9}+\frac{6\pi}{9}=\frac{7\pi}{9}\in]-\pi ;\pi]\)

  • \(k=2\)

    \(x=\frac{2\pi}{9}+\frac{4\pi}{3}=\frac{2\pi}{9}+\frac{12\pi}{9}=\frac{14\pi}{9}\notin]-\pi ;\pi]\)

    \(x=-\frac{2\pi}{9}+\frac{4\pi}{3}=-\frac{2\pi}{9}+\frac{12\pi}{9}=\frac{10\pi}{9}\notin]-\pi ;\pi]\)

    Il est donc inutile de prendre des valeurs de \(k\) plus grandes.

  • \(k=-1\)

    \(x=\frac{2\pi}{9}+\frac{-2\pi}{3}=\frac{2\pi}{9}-\frac{6\pi}{9}=-\frac{-4\pi}{9}\in]-\pi ;\pi]\)

    \(x=-\frac{2\pi}{9}+\frac{-2\pi}{3}=-\frac{2\pi}{9}-\frac{6\pi}{9}=\frac{-8\pi}{9}\in]-\pi ;\pi]\)

  • \(k=-2\)

    \(x=\frac{2\pi}{9}+\frac{-4\pi}{3}=\frac{2\pi}{9}+\frac{-12\pi}{9}=\frac{-10\pi}{9}\notin]-\pi ;\pi]\)

    \(x=-\frac{2\pi}{9}+\frac{-4\pi}{2}=-\frac{2\pi}{9}+\frac{-12\pi}{9}=\frac{-14\pi}{9}\notin]-\pi ;\pi]\)

    Il est donc inutile de prendre des valeurs de \(k\) plus petites.

Question

\(3.sin(4x)= \frac{\sqrt{2}}{2} \quad x\in]-\pi ;\pi]\)

Solution

\(sin(4x)=sin(\frac{\pi}{4})\)

\(\begin{cases}4x=\frac{\pi}{4}+2k\pi\\4x=\pi-frac{\pi}{4}+2k\pi\end{cases}\)

\(\iff \begin{cases}4x=\frac{\pi}{4}+2k\pi\\4x=\frac{\pi}-frac{\pi}{4}+2k\pi\end{cases}\)

\(\iff \begin{cases}4x=\frac{\pi}{4}+2k\pi\\4x=frac{3\pi}{4}+2k\pi\end{cases}\)

\(\iff \begin{cases}x=\frac{\pi}{16}+\frac{k\pi}{2}\\x=frac{3\pi}{16}+\frac{k\pi}{2}\end{cases}\)

  • \(k=0\)

    \(x=\frac{\pi}{16} \in]-\pi ;\pi]\)

    \(x=\frac{3\pi}{16}\in]-\pi ;\pi]\)

  • \(k=1\)

    \(x=\frac{\pi}{16}+\frac{\pi}{2}=\frac{\pi}{16}+\frac{8\pi}{16}=\frac{9\pi}{16}\in]-\pi ;\pi]\)

    \(x=\frac{3\pi}{16}+\frac{\pi}{2}=\frac{3\pi}{16}+\frac{8\pi}{16}=\frac{11\pi}{16}\in]-\pi ;\pi]\)

  • \(k=2\)

    \(x=\frac{\pi}{16}+\frac{2\pi}{2}=\frac{\pi}{16}+\frac{16\pi}{16}=\frac{17\pi}{16}\notin]-\pi ;\pi]\)

    \(x=\frac{3\pi}{16}+\frac{2\pi}{2}=\frac{3\pi}{16}+\frac{16\pi}{16}=\frac{19\pi}{16}\notin]-\pi ;\pi]\)

    Il est donc inutile de prendre des valeurs de \(k\) plus grandes.

  • \(k=-1\)

    \(x=\frac{\pi}{16}+\frac{-\pi}{2}=\frac{\pi}{16}-\frac{8\pi}{16}=-\frac{7\pi}{16}\in]-\pi ;\pi]\)

    \(x=\frac{3\pi}{16}+\frac{-\pi}{2}=\frac{3\pi}{16}+\frac{-8\pi}{16}=\frac{-5\pi}{16}\in]-\pi ;\pi]\)

  • \(k=-2\)

    \(x=\frac{\pi}{16}+\frac{-2\pi}{2}=\frac{\pi}{16}+\frac{-16\pi}{16}=\frac{-15\pi}{16}\in]-\pi ;\pi]\)

    \(x=\frac{3\pi}{16}+\frac{-2\pi}{2}=\frac{3\pi}{16}+\frac{-16\pi}{16}=\frac{-13\pi}{16}\in]-\pi ;\pi]\)

  • \(k=-3\)

    \(x=\frac{\pi}{16}+\frac{-3\pi}{2}=\frac{\pi}{16}+\frac{-24\pi}{16}=\frac{-23\pi}{16}\notin]-\pi ;\pi]\)

    \(x=\frac{3\pi}{16}+\frac{-3\pi}{2}=\frac{3\pi}{16}+\frac{-24\pi}{16}=\frac{-21\pi}{16}\notin]-\pi ;\pi]\)

    Il est donc inutile de prendre des valeurs de \(k\) plus petites.

Question

4. \(sin(2x - \frac{\pi}{3})=sin(3x+\frac{\pi}{2}) \quad x\in]-\pi ;\pi]\)

Solution

\(\begin{cases}2x - \frac{\pi}{3}=3x+\frac{\pi}{2}+2k\pi\\2x - \frac{\pi}{3}=\pi-(3x+\frac{\pi}{2})+2k\pi\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}2x - 3x-\frac{\pi}{3}=\frac{\pi}{2}+2k\pi\\2x - \frac{\pi}{3}=\pi-3x-\frac{\pi}{2}+2k\pi\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}2x - 3x=\frac{\pi}{3}+\frac{\pi}{2}+2k\pi\\2x +3x - \frac{\pi}{3}=\pi-\frac{\pi}{2}+2k\pi\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}-x=\frac{\pi}{3}+\frac{\pi}{2}+2k\pi\\5x=  \frac{\pi}{3} +\pi-\frac{\pi}{2}+2k\pi\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}-x=\frac{2\pi}{6}+\frac{3\pi}{6}+2k\pi\\5x=  \frac{\pi}{3} +\frac{\pi}{2}+2k\pi\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}-x=\frac{5\pi}{6}+2k\pi\\5x=  \frac{2\pi}{6} +\frac{3\pi}{6}+2k\pi\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}-x=\frac{5\pi}{6}+2k\pi\\5x=  \frac{5\pi}{6} +2k\pi\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}x=-\frac{5\pi}{6}-2k\pi\\x=  \frac{\pi}{6} +\frac{2k\pi}{5}\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}x=-\frac{5\pi}{6}-2k\pi\\x=  \frac{\pi}{6} +\frac{2k\pi}{5}\\\end{cases}\) \(k\in\mathbb{Z}\)

  • \(k=0 \):

    \(x=-\frac{5\pi}{6}\in]-\pi ;\pi]\);\(x=  \frac{\pi}{6}\in]-\pi ;\pi]\)

  • \(k=1 \):

    \(x=-\frac{5\pi}{6}-2\pi=-\frac{5\pi}{6}-\frac{12\pi}{6}=-\frac{17\pi}{6}\notin]-\pi ;\pi]\)

    Inutile de prendre des valeurs supérieures de \(k\) pour cette première solution.

    \(x=  \frac{\pi}{6} +\frac{2\pi}{5}\)

    \(\iff x=  \frac{5\pi}{30} +\frac{12\pi}{30}\)

    \(\iff x=  \frac{17\pi}{30}\in]-\pi ;\pi]\)

  • \(k=2 \):

    \(x=  \frac{\pi}{6} +\frac{2\times 2\pi}{5}\)

    \(\iff x=  \frac{\pi}{6} +\frac{4\pi}{5}\)

    \(\iff x=  \frac{5\pi}{30}+\frac{24\pi}{30}\)

    \(\iff x=  \frac{29\pi}{30}\)

  • \(k=3 \):

    \(x=  \frac{\pi}{6} +\frac{3\times 2\pi}{5}\)

    \(\iff x=  \frac{\pi}{6} +\frac{6\pi}{5}\)

    \(\iff x=  \frac{5\pi}{30}+\frac{36\pi}{30}\)

    \(\iff x=  \frac{41\pi}{30}\notin]-\pi ;\pi]\)

    Inutile de prendre des valeurs supérieures de \(k\) pour cette deuxième solution.

  • \(k=-1\)

    \(x=-\frac{5\pi}{6}+2\pi=-\frac{5\pi}{6}+\frac{12\pi}{6}=\frac{7\pi}{6}\notin]-\pi ;\pi]\)

    Inutile de prendre des valeurs inférieures de \(k\) pour cette première solution.

    \(x=  \frac{\pi}{6} +\frac{-1\times 2\pi}{5}\)

    \(\iff x=  \frac{\pi}{6} +\frac{-2\pi}{5}\)

    \(\iff x=  \frac{5\pi}{30}+\frac{-12\pi}{30}\)

    \(\iff x=  \frac{-7\pi}{30}\in]-\pi ;\pi]\)

  • \(k=-2\)

    \(x=  \frac{\pi}{6} +\frac{-2\times 2\pi}{5}\)

    \(\iff x=  \frac{\pi}{6} +\frac{-4\pi}{5}\)

    \(\iff x=  \frac{5\pi}{30}+\frac{-24\pi}{30}\)

    \(\iff x=  \frac{-19\pi}{30}\in]-\pi ;\pi]\)

  • \(k=-3\)

    \(x=  \frac{\pi}{6} +\frac{-3\times 2\pi}{5}\)

    \(\iff x=  \frac{\pi}{6} +\frac{-6\pi}{5}\)

    \(\iff x=  \frac{5\pi}{30}+\frac{-36\pi}{30}\)

    \(\iff x=  \frac{-31\pi}{30}\notin]-\pi ;\pi]\)

    Inutile de prendre des valeurs inférieurs de \(k\) pour cette deuxième solution.

Les solutions de l'équation sont donc :

\(-\frac{5\pi}{6}=-\frac{25\pi}{30} ; \frac{-19\pi}{30} ; \frac{-7\pi}{30} ;\frac{\pi}{6}=\frac{5\pi}{30} ;\frac{17\pi}{30} ;\frac{29\pi}{30}\)

Question

\(5. cos(2x - \frac{\pi}{3})=cos(3x+\frac{\pi}{2}) \quad x\in]-\pi ;\pi]\)

Solution

\(\begin{cases}2x - \frac{\pi}{3}=3x+\frac{\pi}{2}+2k\pi\\2x - \frac{\pi}{3}=-(3x+\frac{\pi}{2})+2k\pi\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}2x - 3x-\frac{\pi}{3}=\frac{\pi}{2}+2k\pi\\2x - \frac{\pi}{3}=-3x-\frac{\pi}{2}+2k\pi\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}2x - 3x=\frac{\pi}{3}+\frac{\pi}{2}+2k\pi\\2x +3x - \frac{\pi}{3}=-\frac{\pi}{2}+2k\pi\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}-x=\frac{\pi}{3}+\frac{\pi}{2}+2k\pi\\5x=  \frac{\pi}{3} -\frac{\pi}{2}+2k\pi\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}-x=\frac{2\pi}{6}+\frac{3\pi}{6}+2k\pi\\5x=  \frac{\pi}{3} -\frac{\pi}{2}+2k\pi\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}-x=\frac{5\pi}{6}+2k\pi\\5x=  \frac{2\pi}{6} -\frac{3\pi}{6}+2k\pi\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}-x=\frac{5\pi}{6}+2k\pi\\5x=  \frac{-\pi}{6} +2k\pi\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}x=-\frac{5\pi}{6}-2k\pi\\x=  \frac{-\pi}{30} +\frac{2k\pi}{5}\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}x=-\frac{5\pi}{6}-2k\pi\\x=  \frac{-\pi}{30} +\frac{2k\pi}{5}\\\end{cases}\) \(k\in\mathbb{Z}\)

  • \(k=0 \):

    \(x=-\frac{5\pi}{6}\in]-\pi ;\pi]\);\(x=  \frac{-\pi}{30}\in]-\pi ;\pi]\)

  • \(k=1 \):

    \(x=-\frac{5\pi}{6}-2\pi=-\frac{5\pi}{6}-\frac{12\pi}{6}=-\frac{17\pi}{6}\notin]-\pi ;\pi]\)

    Inutile de prendre des valeurs supérieures de \(k\) pour cette première solution.

    \(x=  \frac{-\pi}{30} +\frac{2\pi}{5}\)

    \(\iff x=  \frac{-\pi}{30} +\frac{12\pi}{30}\)

    \(\iff x=  \frac{11\pi}{30}\in]-\pi ;\pi]\)

  • \(k=2 \):

    \(x=  \frac{-\pi}{30} +\frac{2\times 2\pi}{5}\)

    \(\iff x=  \frac{-\pi}{30} +\frac{4\pi}{5}\)

    \(\iff x=  \frac{-\pi}{30}+\frac{24\pi}{30}\)

    \(\iff x=  \frac{23\pi}{30}\)

  • \(k=3 \):

    \(x=  \frac{-\pi}{30} +\frac{3\times 2\pi}{5}\)

    \(\iff x=  \frac{-\pi}{30} +\frac{6\pi}{5}\)

    \(\iff x=  \frac{-\pi}{30}+\frac{36\pi}{30}\)

    \(\iff x=  \frac{35\pi}{30}\notin]-\pi ;\pi]\)

    Inutile de prendre des valeurs supérieures de \(k\) pour cette deuxième solution.

  • \(k=-1\)

    \(x=-\frac{5\pi}{6}+2\pi=-\frac{5\pi}{6}+\frac{12\pi}{6}=\frac{7\pi}{6}\notin]-\pi ;\pi]\)

    Inutile de prendre des valeurs inférieures de \(k\) pour cette première solution.

    \(x=  \frac{-\pi}{30} +\frac{-1\times 2\pi}{5}\)

    \(\iff x=  \frac{-\pi}{30} +\frac{-2\pi}{5}\)

    \(\iff x=  \frac{-\pi}{30}+\frac{-12\pi}{30}\)

    \(\iff x=  \frac{-13\pi}{30}\in]-\pi ;\pi]\)

  • \(k=-2\)

    \(x=  \frac{-\pi}{30} +\frac{-2\times 2\pi}{5}\)

    \(\iff x=  \frac{-\pi}{30} +\frac{-4\pi}{5}\)

    \(\iff x=  \frac{-\pi}{30}+\frac{-24\pi}{30}\)

    \(\iff x=  \frac{-25\pi}{30}\in]-\pi ;\pi]\)

  • \(k=-3\)

    \(x=  \frac{-\pi}{30} +\frac{-3\times 2\pi}{5}\)

    \(\iff x=  \frac{-\pi}{30} +\frac{-6\pi}{5}\)

    \(\iff x=  \frac{-\pi}{30}+\frac{-36\pi}{30}\)

    \(\iff x=  \frac{-37\pi}{30}\notin]-\pi ;\pi]\)

    Inutile de prendre des valeurs inférieurs de \(k\) pour cette deuxième solution.

Les solutions de l'équation sont donc :

\(-\frac{5\pi}{6}=-\frac{25\pi}{30}\);\(\frac{-13\pi}{30}\in]-\pi ;\pi]\);\(\frac{-\pi}{30}\);

\(\frac{11\pi}{30}\);\(\frac{23\pi}{30}\)

Question

\(6. cos(2x - \frac{\pi}{3})=sin(3x+\frac{\pi}{2}) \quad x\in]-\pi ;\pi]\)

Solution

\(sin(x)=cos(\frac{\pi}{2}-x)\) pour tout \(x\in \mathbb{R}\)

\(cos(2x - \frac{\pi}{3})=cos(\frac{\pi}{2}-(3x+\frac{\pi}{2}))\)

\(\iff cos(2x - \frac{\pi}{3})=cos(\frac{\pi}{2}-3x-\frac{\pi}{2})\)

\(\iff cos(2x - \frac{\pi}{3})=cos(-3x)\)

\(cos(-x)=cos(x)\) pour tout \(x\in \mathbb{R}\)

\(\iff cos(2x - \frac{\pi}{3})=cos(3x)\)

\(\begin{cases}2x - \frac{\pi}{3}=3x+2k\pi\\2x - \frac{\pi}{3}=-3x+2k\pi\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}2x - 3x-\frac{\pi}{3}=2k\pi\\2x - \frac{\pi}{3}=-3x+2k\pi\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}2x - 3x=\frac{\pi}{3}+2k\pi\\2x +3x - \frac{\pi}{3}=2k\pi\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}-x=\frac{\pi}{3}+2k\pi\\5x=  \frac{\pi}{3} +2k\pi\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}x=-\frac{\pi}{3}-2k\pi\\x= \frac{\pi}{15} +\frac{2k\pi}{5}\\\end{cases}\) \(k\in\mathbb{Z}\)

  • \(k=0 \):

    \(x=-\frac{\pi}{3}\in]-\pi ;\pi]\);\(x=  \frac{\pi}{15}\in]-\pi ;\pi]\)

  • \(k=1 \):

    \(x=-\frac{\pi}{3}-2\pi=-\frac{\pi}{3}-\frac{6\pi}{3}=-\frac{7\pi}{3}\notin]-\pi ;\pi]\)

    Inutile de prendre des valeurs supérieures de \(k\) pour cette première solution.

    \(x=  \frac{\pi}{15} +\frac{\pi}{5}\)

    \(\iff x=  \frac{\pi}{15} +\frac{6\pi}{15}\)

    \(\iff x=  \frac{7\pi}{15}\in]-\pi ;\pi]\)

  • \(k=2 \):

    \(x=  \frac{\pi}{15} +\frac{2\times 2\pi}{5}\)

    \(\iff x=  \frac{\pi}{15} +\frac{4\pi}{5}\)

    \(\iff x=  \frac{\pi}{15}+\frac{12\pi}{15}\)

    \(\iff x=  \frac{13\pi}{15}\)

  • \(k=3 \):

    \(x=  \frac{\pi}{15} +\frac{3\times 2\pi}{5}\)

    \(\iff x=  \frac{\pi}{15} +\frac{6\pi}{5}\)

    \(\iff x=  \frac{\pi}{15}+\frac{18\pi}{15}\)

    \(\iff x=  \frac{19\pi}{15}\notin]-\pi ;\pi]\)

    Inutile de prendre des valeurs supérieures de \(k\) pour cette deuxième solution.

  • \(k=-1\)

    \(x=-\frac{\pi}{3}+2\pi=-\frac{\pi}{3}+\frac{6\pi}{3}=\frac{5\pi}{3}\notin]-\pi ;\pi]\)

    Inutile de prendre des valeurs inférieures de \(k\) pour cette première solution.

    \(x=  \frac{\pi}{15} +\frac{-1\times 2\pi}{5}\)

    \(\iff x=  \frac{\pi}{15} +\frac{-2\pi}{5}\)

    \(\iff x=  \frac{\pi}{15}+\frac{-6\pi}{15}\)

    \(\iff x=  \frac{-5\pi}{15}\in]-\pi ;\pi]\)

  • \(k=-2\)

    \(x=  \frac{\pi}{15} +\frac{-2\times 2\pi}{5}\)

    \(\iff x=  \frac{\pi}{15} +\frac{-4\pi}{5}\)

    \(\iff x=  \frac{\pi}{15}+\frac{-12\pi}{15}\)

    \(\iff x=  \frac{-11\pi}{15}\in]-\pi ;\pi]\)

  • \(k=-3\)

    \(x=  \frac{\pi}{15} +\frac{-3\times 2\pi}{5}\)

    \(\iff x=  \frac{\pi}{15} +\frac{-6\pi}{5}\)

    \(\iff x=  \frac{\pi}{15}+\frac{-36\pi}{30}\)

    \(\iff x=  \frac{-35\pi}{15}\notin]-\pi ;\pi]\)

    Inutile de prendre des valeurs inférieurs de \(k\) pour cette deuxième solution.

Les solutions de l'équation sont donc :

\(\frac{-11\pi}{15} \);\(-\frac{\pi}{3}\)=\(-\frac{5\pi}{15}\);\(\frac{\pi}{15}\);\(\frac{7\pi}{15}\);\(\frac{13\pi}{15}\)

Question

\(7. sin(2x - \frac{\pi}{3})=cos(3x+\frac{\pi}{2}) \quad x\in]-\pi ;\pi]\)

Solution

\(sin(x)=cos(\frac{\pi}{2}-x)\) pour tout \(x\in \mathbb{R}\)

\(sin(2x - \frac{\pi}{3})=cos(3x+\frac{\pi}{2})\)

\(\iff cos(\frac{\pi}{2}-(2x - \frac{\pi}{3}))=cos(3x+\frac{\pi}{2})\)

\(\iff cos(\frac{\pi}{2}-(2x - \frac{\pi}{3}))=cos(3x+\frac{\pi}{2})\)

\(\iff cos(\frac{\pi}{2}-2x + \frac{\pi}{3})=cos(3x+\frac{\pi}{2})\)

\(\iff cos(\frac{3\pi}{6}-2x + \frac{2\pi}{6})=cos(3x+\frac{\pi}{2})\)

\(\iff cos(-2x+\frac{5\pi}{6})=cos(3x+\frac{\pi}{2})\)

\(\begin{cases}-2x+\frac{5\pi}{6}=3x+\frac{\pi}{2}+2k\pi\\-2x+\frac{5\pi}{6}=-(3x+\frac{\pi}{2})+2k\pi\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}-2x - 3x+\frac{5\pi}{6}=\frac{\pi}{2}+2k\pi\\-2x + \frac{5\pi}{6}=-3x-\frac{\pi}{2}+2k\pi\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}-2x - 3x=\frac{\pi}{2}-\frac{5\pi}{6}+2k\pi\\-2x +3x + \frac{5\pi}{6}=-\frac{\pi}{2}+2k\pi\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}-5x=\frac{\pi}{2}-\frac{5\pi}{6}+2k\pi\\x=  -\frac{\pi}{2}-\frac{5\pi}{6} +2k\pi\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}-5x=\frac{3\pi}{6}-\frac{5\pi}{6}+2k\pi\\x= -\frac{3\pi}{6} -\frac{5\pi}{6}+2k\pi\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}-5x=\frac{-2\pi}{6}+2k\pi\\x= -\frac{8\pi}{6} +2k\pi\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}5x=\frac{\pi}{3}-2k\pi\\x= -\frac{4\pi}{3} +2k\pi\\\end{cases}\) \(k\in\mathbb{Z}\)

\(\iff \begin{cases}x=\frac{\pi}{15}-\frac{2k\pi}{5}\\x= -\frac{4\pi}{3} +2k\pi\\\end{cases}\) \(k\in\mathbb{Z}\)

  • \(k=0 \):

    \(x=\frac{\pi}{15}\in]-\pi ;\pi]\);\(x=  -\frac{4\pi}{3}\notin]-\pi ;\pi]\)

  • \(k=1 \):

    \(x=\frac{\pi}{15}-\frac{2k\pi}{5}=\frac{\pi}{15}-\frac{6\pi}{15}=-\frac{5\pi}{15}\in]-\pi ;\pi]\)

    \(x=  \frac{-4\pi}{3} +2\pi\)

    \(\iff x=  \frac{-4\pi}{3} +\frac{6\pi}{3}\)

    \(\iff x=  \frac{2\pi}{3}\in]-\pi ;\pi]\)

  • \(k=2 \):

    \(x=  \frac{\pi}{15} -\frac{2\times 2\pi}{5}\)

    \(\iff x=  \frac{\pi}{15} -\frac{4\pi}{5}\)

    \(\iff x=  \frac{\pi}{15}-\frac{12\pi}{15}\)

    \(\iff x=  \frac{-11\pi}{15}\in]-\pi ;\pi]\)

    \(x=  \frac{-4\pi}{3} +2 \times 2\pi\)

    \(\iff x=  \frac{-4\pi}{3} +\frac{12\pi}{3}\)

    \(\iff x=  \frac{8\pi}{3}\notin]-\pi ;\pi]\)

    Inutile de prendre des valeurs de \(k\) plus grande pour cette

    deuxième solution.

  • \(k=3 \):

    \(x=  \frac{\pi}{15} -\frac{3\times 2\pi}{5}\)

    \(\iff x=  \frac{\pi}{15} -\frac{6\pi}{5}\)

    \(\iff x=  \frac{\pi}{15}-\frac{18\pi}{15}\)

    \(\iff x=  -\frac{17\pi}{15}\notin]-\pi ;\pi]\)

    Inutile de prendre des valeurs supérieures de \(k\) pour cette deuxième solution.

  • \(k=-1\)

    \(x=\frac{\pi}{15}+\frac{2\pi}{5}=\frac{\pi}{15}+\frac{6\pi}{15}\)

    \(\iff x=\frac{7\pi}{15}\in]-\pi ;\pi]\)

    \(x=  \frac{-4\pi}{3} -1 \times 2\pi\)

    \(\iff x=  \frac{-4\pi}{3} - \frac{6\pi}{3}\)

    \(\iff x=  \frac{-10\pi}{3}\notin]-\pi ;\pi]\)

    Inutile de prendre des valeurs inférieures de \(k\) pour cette deuxième solution.

  • \(k=-2\)

    \(x=\frac{\pi}{15}+\frac{2 \times 2\pi}{5}\)

    \(\iff x=\frac{\pi}{15}+\frac{4\pi}{5}\)

    \(\iff x=\frac{\pi}{15}+\frac{12\pi}{15}\)

    \(\iff x=\frac{13\pi}{15}\in]-\pi ;\pi]\)

  • \(k=-3\)

    \(x=\frac{\pi}{15}+\frac{3 \times 2\pi}{5}\)

    \(\iff x=\frac{\pi}{15}+\frac{6\pi}{5}\)

    \(\iff x=\frac{\pi}{15}+\frac{18\pi}{15}\)

    \(\iff x=\frac{19\pi}{15}\notin]-\pi ;\pi]\)

    Inutile de prendre des valeurs inférieures de \(k\) inférieur pour cette première équation.

Les solutions de l'équation sont donc :

\(-\frac{11\pi}{15} ;-\frac{5\pi}{15};{\pi}{15};\frac{7\pi}{15};\frac{2\pi}{3}=\frac{10\pi}{15};\frac{13\pi}{15}\)

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