$$\cos(2x)-\cos(3x)=0$$
I am trying to solve this equation but get stuck every time in the middle of the exercise. Can somebody help me please?
$\begingroup$
$\endgroup$
1
-
$\begingroup$ But this isn't a site to ask such problems... $\endgroup$ – Aditya Sep 27 '17 at 16:02
Add a comment
|
$\begingroup$
$\endgroup$
8
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.
-
$\begingroup$ Unexpected donwvote for the best answer... $\endgroup$ – Yves Daoust Sep 27 '17 at 15:56
-
-
1
-
$\begingroup$ @velutluna Did you miss the "or $3x=2n\pi-2x$" ? $\endgroup$ – user228113 Sep 27 '17 at 15:58
-
$\begingroup$
$\endgroup$
0
Hint: $$\cos 2x - \cos 3x = 2 \sin \frac{5x}{2}\sin\frac{x}{2}$$
$\begingroup$
$\endgroup$
use that $$\cos(x)-\cos(y)=-2 \sin \left(\frac{x}{2}-\frac{y}{2}\right) \sin \left(\frac{x}{2}+\frac{y}{2}\right)$$
