Find the equation whose roots are equal to numbers $$\sin^2\frac{\pi}{2n + 1},\ \sin^2\frac{2\pi}{2n + 1},\ \sin^2\frac{3\pi}{2n + 1},\dots,\ \sin^2\frac{n\pi}{2n + 1}$$
The question looked intriguing and now with the hints provided here I can take it forward.
On
Hint:
For $n = 1$ your numbers are
$$\sin^2 {\pi \over 3},\ \sin^2 {2\pi \over 3},\ \sin^2 {3\pi \over 3},\ $$
whereas $$\sin^2 {\pi \over 3}=\sin^2 {2\pi \over 3},$$
The similar situation is for all $n$: $n$ pairs of equal values:
Now, all red points $z = \cos \phi + \mathbb i\sin \phi$ represent all solutions of the equation $$z^{2n+1} = -1,$$ i.e. $$(\cos \phi + \mathbb i\sin \phi)^{2n+1} = -1$$
Consider to expand the left-hand side and then separate real and imaginary parts of the equation.
Maybe you should check the original question... It was probably not the one you are asking. One possible (and trivial) answer to your question can just be $$ (z-\sin^2\frac{\pi}{2n+1})(z-\sin^2\frac{2\pi}{2n+1})\cdots(z-\sin^2\frac{n\pi}{2n+1})=0 $$