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$$\cos(2x)-\cos(3x)=0$$

I am trying to solve this equation but get stuck every time in the middle of the exercise. Can somebody help me please?

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  • $\begingroup$ But this isn't a site to ask such problems... $\endgroup$ – Aditya Sep 27 '17 at 16:02
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Notice, we have $$\cos(3x)=\cos(2x)$$ $$3x=2n\pi\pm 2x$$ Where, $n$ is any integer

Now, we have the following solutions

$$3x=2n\pi+2x\implies \color{red}{x=2n\pi}$$ or $$3x=2n\pi-2x\implies \color{red}{x=\frac{2n\pi}{5}}$$

Edit -1 (Thanks @Bernard)

Observe the second set of solutions contains the first.

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  • $\begingroup$ Unexpected donwvote for the best answer... $\endgroup$ – Yves Daoust Sep 27 '17 at 15:56
  • $\begingroup$ I was waiting for that $\endgroup$ – Aditya Sep 27 '17 at 15:57
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    $\begingroup$ @velutluna: yes it does, with a $\pm$. $\endgroup$ – Yves Daoust Sep 27 '17 at 15:58
  • $\begingroup$ @velutluna Did you miss the "or $3x=2n\pi-2x$" ? $\endgroup$ – user228113 Sep 27 '17 at 15:58
  • $\begingroup$ I see! My mistake! $\endgroup$ – velut luna Sep 27 '17 at 16:00
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Hint: $$\cos 2x - \cos 3x = 2 \sin \frac{5x}{2}\sin\frac{x}{2}$$

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use that $$\cos(x)-\cos(y)=-2 \sin \left(\frac{x}{2}-\frac{y}{2}\right) \sin \left(\frac{x}{2}+\frac{y}{2}\right)$$

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