# Trigonometry equation $\cos(2x)-\cos(3x)=0$

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

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

• But this isn't a site to ask such problems... – Aditya Sep 27 '17 at 16:02

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.

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

Hint: $$\cos 2x - \cos 3x = 2 \sin \frac{5x}{2}\sin\frac{x}{2}$$

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