resoudre dans $$ \mathbb{R} $$ $$ 2 \cos x \sin x=\cos 3 x $$

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Solve the equation $$ 2 \cos x \sin x=\cos 3 x $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$2\cos\left(x\right)\sin\left(x\right)=\cos\left(3x\right)$$
- step1: Calculate:
  $$\sin\left(2x\right)=\cos\left(3x\right)$$
- step2: Rewrite the expression:
  $$2\sin\left(x\right)\cos\left(x\right)=4\cos^{3}\left(x\right)-3\cos\left(x\right)$$
- step3: Move the expression to the left side:
  $$2\sin\left(x\right)\cos\left(x\right)-\left(4\cos^{3}\left(x\right)-3\cos\left(x\right)\right)=0$$
- step4: Calculate:
  $$2\sin\left(x\right)\cos\left(x\right)-4\cos^{3}\left(x\right)+3\cos\left(x\right)=0$$
- step5: Factor the expression:
  $$\cos\left(x\right)\left(2\sin\left(x\right)-4\cos^{2}\left(x\right)+3\right)=0$$
- step6: Separate into possible cases:
  $$\begin{align}&\cos\left(x\right)=0\&2\sin\left(x\right)-4\cos^{2}\left(x\right)+3=0\end{align}$$
- step7: Solve the equation:
  $$\begin{align}&x=\frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\&x=\left\{ \begin{array}{l}\frac{\pi }{10}+2k\pi \\frac{9\pi }{10}+2k\pi \\frac{13\pi }{10}+2k\pi \\frac{17\pi }{10}+2k\pi \end{array}\right.,k \in \mathbb{Z}\end{align}$$
- step8: Find the union:
  $$x=\left\{ \begin{array}{l}\frac{\pi }{10}+2k\pi \\frac{\pi }{2}+k\pi \\frac{9\pi }{10}+2k\pi \\frac{13\pi }{10}+2k\pi \\frac{17\pi }{10}+2k\pi \end{array}\right.,k \in \mathbb{Z}$$
La solution de l'équation $$2 \cos x \sin x = \cos 3x$$ dans $$\mathbb{R}$$ est donnée par :

$$x = \left\{ \begin{array}{l} \frac{\pi}{10} + 2k\pi \ \frac{\pi}{2} + k\pi \ \frac{9\pi}{10} + 2k\pi \ \frac{13\pi}{10} + 2k\pi \ \frac{17\pi}{10} + 2k\pi \end{array} \right., k \in \mathbb{Z}$$

où $$k$$ est un entier.
Les solutions de l'équation $$2 \cos x \sin x = \cos 3x$$ dans $$\mathbb{R}$$ sont : $$x = \frac{\pi}{10} + 2k\pi,\ \frac{\pi}{2} + k\pi,\ \frac{9\pi}{10} + 2k\pi,\ \frac{13\pi}{10} + 2k\pi,\ \frac{17\pi}{10} + 2k\pi$$ où $$k$$ est un entier.

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