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.
 A: 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.
A: The following figure shows that the $2\pi$-periodic functions $x\mapsto \cos x$, $\,\cos(2x)$, and $\cos(3x)$ in red, and the functions $x\mapsto\sin x$, $\,\sin(2x)$, and $\sin(3x)$ in blue. We have to look at the red-blue crossings. Three such crossings at the same $x$ constitute a hit. 

Inspecting the figure shows that we could have such hits at $x={\pi\over8}$ and at $x=-{3\pi\over8}$. Inserting these $x$ into the set equation
$$\{\sin x,\sin(2x),\sin(3x)\}=\{\cos x,\cos(2x),\cos(3x)\}\tag{1}$$
shows that they indeed solve the equation. Due to symmetry is not necessary to compute $\cos{\pi\over8}$ and similars. We have
$$\sin{\pi\over8}=\sin\left({\pi\over4}-{\pi\over8}\right)=\cos\left({\pi\over4}+{\pi\over8}\right)=\cos{3\pi\over8}\ ,$$and similar. Similarly, when $x=-{3\pi\over8}$.
In order to show that there are no other solutions it is sufficient to "intersect" $x\mapsto\sin x$ with $x\mapsto \cos x$, $\,\cos(2x)$, and $\cos(3x)$ and to verify that the points of intersection lead to no other full solutions of the problem. I have not found a simpler way.
A: For the sets to be the same you need the sums to be identical but also all symmetric sum-product combinations must be identical.  Thus
$\sin x+\sin 2x+\sin 3x= \cos x+\cos 2x +\cos 3x$
$\sin x\sin 2x +\sin x\sin 3x+\sin 2x\sin 3x=\cos x\cos 2x+\cos x\cos 3x+\cos 2x\cos 3x$
$\sin x\sin 2x\sin 3x=\cos x\cos 2x\cos 3x$
Iff all three of these relations are satisfied for some $x$ then the sets are identical.  You can get candidates by solving the first relation but then you have to check them against the other two.
To simplify this checking process you can simplify the products by using the sum-product relations, for instance
$\sin x\sin 2x= (1/2)(\cos x-\cos 3x)$
$\cos x\cos 2x= (1/2)(\cos x+\cos 3x)$
When you apply these sum-product relations and do some cancellations you get the following for the second and third equations which you will use for checking:
$\cos 3x +\cos 4x+\cos 5x=0$ from the second eqn.
$\sin 2x+\sin 4x-\sin 6x=1+\cos 2x+\cos 4x+\cos 6x$ from the third eqn.
With these formulations the relatively cumbersome functions involving $\pi/8$ and $3\pi/8$ cancel when you check that candidate.
