Find solution for a trigonometric equation set How can I solve
$$\begin{cases}
\sin^2(y_1) = \frac{1}{2}\sin^2(\frac{y_1+y_2}{2})\\
\sin^2(y_2) = \frac{1}{2}(\sin^2(\frac{y_1+y_2}{2})+1)
\end{cases}$$
where $$\space y_1,y_2 \in [0,\frac{\pi}{2}] $$
Thanks a lot!
 A: Graphing seems to show $16$ solutions in $0 \le y_1, y_2 \le 2\pi$, but only one with $0 \le y_1, y_2 \le \pi/2$.

Letting $z_i = \exp(i y_i/2)$ and expanding everything out, we get the polynomial equations
$$\eqalign{ -2\,{z_{{1}}}^{8}{z_{{2}}}^{2}+{z_{{1}}}^{6}{z_{{2}}}^{4}+2\,{z_{{1}}
}^{4}{z_{{2}}}^{2}+{z_{{1}}}^{2}-2\,{z_{{2}}}^{2}&=0\cr{z_{{1}}}^{4}{z_{{2}
}}^{6}-2\,{z_{{1}}}^{2}{z_{{2}}}^{8}-2\,{z_{{1}}}^{2}{z_{{2}}}^{4}-2\,
{z_{{1}}}^{2}+{z_{{2}}}^{2}&=0}
$$
The resultant of these two polynomials with respect to $z_2$ is
$$ 256\,{z_{{1}}}^{8} \left( 3\,{z_{{1}}}^{16}-6\,{z_{{1}}}^{12}+7\,{z_{{
1}}}^{8}-6\,{z_{{1}}}^{4}+3 \right) ^{2} \left( {z_{{1}}}^{8}-{z_{{1}}
}^{4}+1 \right) ^{4}
$$
It turns out that the solution we want has $z_1^8 - z_1^4 + 1 = 0$: this says $z_1^4$ is a cube root of $-1$, and in fact $z_1 = \exp(i \pi/12)$, $z_2 = \exp(i \pi/6)$, so $y_1 = \pi/6$, $y_2 = \pi/3$. 
A: EDITION.- For beginners I add a third figure with the "explanation" of the graphic (with infinitely many closed curves).
COMMENT.- The given system easily leds to the equation $$\sin^2(y)-\sin^2(x)=\frac 14$$
The graphic of this equation is something amazing (see figure below).
On the other hand we have the condition  $\space y_1,y_2 \in [0,\frac{\pi}{2}]$  so we have to see only at the square $[0,\frac{\pi}{2}]\text{ x }[0,\frac{\pi}{2}]$.With appropriate scale on the graph, we can see that we are concerned just with an arc (enclosed by the square) of just one of the closed curves in the figure.The solutions of the given system are in this arc. 



