Solve cos(x) + cos(3x) = sin(x) + sin(3x)

[Alettix]

Homework Statement

The following equation is to be solved for all x:
$\cos(x) + \cos(3x) = \sin(x) + \sin(3x)$

Homework Equations

The tripple angle formulas:
$\cos(3x) = 4\cos^3(x) - 3\cos(x)$
$\sin(3x) = 3\sin(x) - 4\sin^3(x)$
The Pythagorean trig identity:
$\sin^2(x) + \cos^2(x) = 1$

The Attempt at a Solution

Applying the tripple angle identities we have:
$\cos(x) + 4\cos^3(x) - 3\cos(x) = \sin(x) + 3\sin(x) - 4\sin^3(x)$
Simplifying:
$4\cos^3(x) - 2\cos(x) = 4\sin(x) - 4\sin^3(x)$
$2\cos(x)(2\cos^2(x) - 1) = 4\sin(x)(1 - \sin^2(x)$
With the Pythagorean identity:
$\cos(x)(\cos^2(x) - \sin^2(x)) = 2\sin(x)\cos^2(x)$

Now, from this it look as if $cos(x) = 0$ should be a solution, which yields $x_1 = \pi n$ where $n$ is an integer. Continuing with the rest:

$(\cos^2(x) - \sin^2(x)) = 2\sin(x)\cos(x)$
$\cot(x) - \tan(x) = 2$
$\tan^2(x) + 2\tan(x) -1 =0$

Solving the second degree equation yields $\tan(x) = -1 \pm \sqrt{2}$, which gives $x_2 = \frac{\pi}{8} +\pi n$ and $x_3 = \frac{-3\pi}{8} +\pi n$.

Now the only problem is that none of these solutions is right! Here https://www.desmos.com/calculator/dvbz4qpadt I plotted the functions and searched their intersection, and it doesn't match my solution. Where is my misstake? How can I solve the equation properly?

Thank you very much in advance!

 

[micromass]

$cos(x) = 0$ should be a solution, which yields $x_1 = \pi n$ where $n$ is an integer.

Are you sure that $\cos(n\pi) = 0$?

 

[Alettix]

Are you sure that $\cos(n\pi) = 0$?

Oh noo! That's totally wrong! It yields $x_1 = \frac{\pi}{2} + \pi n$ of course!

 

[micromass]

Yep. And now it seems the graph agrees with your solutions!

 

[Alettix]

Yep. And now it seems the graph agrees with your solutions!

And now I see that the solutions are fine...Such a stupid misstake!
Thank you for your help!

 

[haruspex]

Fwiw, you could have gone
$\cos(2x-x)+\cos(2x+x)=\sin(2x-x)+\sin(2x+x)$
$2\cos(2x)\cos(x)=2\sin(2x)\cos(x)$
$\cos(x)=0$ or $\tan(2x)=1$.