Find such $x$ so that set $\{\sin x, \sin2x, \sin3x\}$ coincides with the set $\{\cos x, \cos2x, \cos3x\}$ I recently started learning set theory so my attempt might be looking silly.
I set the sums of elements to be equal:
$$
\begin{split}
\sin x+\sin 2x+\sin 3x&=\cos x+\cos 2x+\cos 3x\\
\sin 2x+2\sin 2x\cos x&=\cos 2x+2\cos 2x \cos x\\
(\sin 2x-\cos 2x)(2\cos x+1)&=0\\
\end{split}
$$


*

*1) 


$\sin2x-\cos2x=0$
$\tan2x=1$
$x=\pi/8+\pi n/2$


*

*2) 


$\cos x=-1/2$
$x=\pm\pi/3+\pi k$
Turns out that the solution of 1) is the correct answer to the initial problem, but the solution of 2) isn't.

Your first equation is necessary, but not sufficient. If $\tan 2x=1$ then $\tan x=\pm\sqrt{2}-1$ and $\tan 3x=\frac{1+\tan x}{1-\tan x}=\frac{\pm\sqrt{2}}{2\mp\sqrt{2}}=\pm\sqrt{2}+1=\frac{1}{\tan x}$, so that works. By contrast, if $\cos x=-\frac12$ then $\cos 2x=2\cos^2x-1=-\frac12$ and $\cos3x=\cos x(2\cos 2x-1)=-1$, so $\sin x=\pm\frac{\sqrt{3}}{2}$ isn't obtained.
