Solving a system of equations which contain sin and cosine terms. Hello my question is the following: Solve the given system of equations:
$$E=\frac{l_{p}}{\pi}\sqrt{\sin^{2}\left(\frac{\pi y_{1}}{l_{p}}\right)+\sin^{2}\left(\frac{\pi y_{2}}{l_{p}}\right)+\sin^{2}\left(\frac{\pi y_{3}}{l_{p}}\right)+\sin^{2}\left(\frac{\pi y_{4}}{l_{p}}\right)};$$
$$\frac{t\cos\left(\frac{\pi y_{1}}{l_{p}}\right)\sin\left(\frac{\pi y_{1}}{l_{p}}\right)}{\sqrt{\sin^{2}\left(\frac{\pi y_{1}}{l_{p}}\right)+\sin^{2}\left(\frac{\pi y_{2}}{l_{p}}\right)+\sin^{2}\left(\frac{\pi y_{3}}{l_{p}}\right)+\sin^{2}\left(\frac{\pi y_{4}}{l_{p}}\right)}}=n_{1}l_{q} ;$$
$$\frac{t\cos\left(\frac{\pi y_{2}}{l_{p}}\right)\sin\left(\frac{\pi y_{2}}{l_{p}}\right)}{\sqrt{\sin^{2}\left(\frac{\pi y_{1}}{l_{p}}\right)+\sin^{2}\left(\frac{\pi y_{2}}{l_{p}}\right)+\sin^{2}\left(\frac{\pi y_{3}}{l_{p}}\right)+\sin^{2}\left(\frac{\pi y_{4}}{l_{p}}\right)}}=n_{2}l_{q} ;$$
$$\frac{t\cos\left(\frac{\pi y_{3}}{l_{p}}\right)\sin\left(\frac{\pi y_{3}}{l_{p}}\right)}{\sqrt{\sin^{2}\left(\frac{\pi y_{1}}{l_{p}}\right)+\sin^{2}\left(\frac{\pi y_{2}}{l_{p}}\right)+\sin^{2}\left(\frac{\pi y_{3}}{l_{p}}\right)+\sin^{2}\left(\frac{\pi y_{4}}{l_{p}}\right)}}=n_{3}l_{q} ;$$
$$\frac{t\cos\left(\frac{\pi y_{4}}{l_{p}}\right)\sin\left(\frac{\pi y_{4}}{l_{p}}\right)}{\sqrt{\sin^{2}\left(\frac{\pi y_{1}}{l_{p}}\right)+\sin^{2}\left(\frac{\pi y_{2}}{l_{p}}\right)+\sin^{2}\left(\frac{\pi y_{3}}{l_{p}}\right)+\sin^{2}\left(\frac{\pi y_{4}}{l_{p}}\right)}}=n_{4}l_{q}; $$
for $y_{1} ,y_{2} ,y_{3} , y_{4},$ and $t$, where $n_{1}, n_{2}, n_{3}, n_{4}, l_{q}, l_{p},$ and $E$ are constants.
I have found two solutions $y_{i}=\frac{l_{p}}{\pi}\arcsin\left(\frac{\pm E\pi}{2l_{p}}\right)$, where $i=1,..,4$. Using 
$$\frac{t\left(\cos\left(\frac{\pi y_{1}}{l_{p}}\right)\sin\left(\frac{\pi y_{1}}{l_{p}}\right)+\cos\left(\frac{\pi y_{2}}{l_{p}}\right)\sin\left(\frac{\pi y_{2}}{l_{p}}\right)+\cos\left(\frac{\pi y_{3}}{l_{p}}\right)\sin\left(\frac{\pi y_{3}}{l_{p}}\right)+\cos\left(\frac{\pi y_{4}}{l_{p}}\right)\sin\left(\frac{\pi y_{4}}{l_{p}}\right)\right)}{\sqrt{\sin^{2}\left(\frac{\pi y_{1}}{l_{p}}\right)+\sin^{2}\left(\frac{\pi y_{2}}{l_{p}}\right)+\sin^{2}\left(\frac{\pi y_{3}}{l_{p}}\right)+\sin^{2}\left(\frac{\pi y_{4}}{l_{p}}\right)}}-l_{q}(n_{1}+n_{2}+n_{3}+n_{4})=0 $$
for $t$, then substituting the values for $y_{i}$ into it, as well as using the $E\pi/l_{p}$ for the value of the square root, gives $t$.
This however is assuming that the $y_{i}$'s are equal, I'm struggling to find points where $y_{1}\neq y_{2}\neq y_{3}\neq y_{4}$. Any suggestions?  I have also tried to solve it numerically, but want I solutions in a general form. I have also tried polar coordinates, but that didn't seem to help... Thanks guys.
 A: Hint:
After some changes of variables and constants (such as $a=2\sin^2(\lambda y_1)-1$), your system can be written as the intersection of four hyperquadrics and a hyperplane.
$$a^2=1-Au,\\
b^2=1-Bu,\\
c^2=1-Cu,\\
d^2=1-Du,\\
a+b+c+d=E.$$
A: Let's work on the first equation:
$$ \dfrac{t\cos{ \left( \frac{\pi y_1}{l_p} \right) } \sin { \left( \frac{\pi y_1}{l_p} \right) } } {\sqrt{ \sin^2 { \left( \frac{\pi y_1}{l_p} \right) } + \sin^2 { \left( \frac{\pi y_1}{l_p} \right) } + \sin^2 { \left( \frac{\pi y_1}{l_p} \right) } +\sin^2 { \left( \frac{\pi y_1}{l_p} \right) } }} = \dfrac{\frac{t}{2} \sin { \left( \frac{2\pi y_1}{l_p} \right) } }{ \frac {E\pi}{l_p}} = n_1l_q$$
This gives the equation:
$$\sin { \left( \frac{2\pi y_1}{l_p} \right) } = \dfrac{2E\pi n_1l_q}{tl_p} $$
I used the fact that the whole denominator is a constant, from what you stated. This should give an equation for each $y_i$.
I hope I understood your question correctly. Can you continue from here? 
Edit: I didn't. I missed out on the fact that $t$ is a variable. I have an idea though.
Look at this:
$$ \cos 2x = 1 - 2\sin^2 x $$
$$ \cos^2 2x = \left( 1 - 2\sin^2 x \right)^2 $$
$$ \sin^2 x  = \dfrac{1-\sqrt{1-\sin^2 2x}}{2}  $$
Threfore, we have:
$$ \sin^2 \left( \frac{\pi y_1}{l_p} \right) = \dfrac{1-\sqrt{1-\sin^2 \left( \frac{2\pi y_1 }{l_p} \right)}}{2} = \dfrac{1-\sqrt{1-\dfrac{4E^2 \pi^2 n_1^2 l_q^2 }{t^2 l_p^2}}}{2}$$
Doing this for every sine and plugging in the first equation gives an equation for $t$:
$$ E = \dfrac{l_p}{\pi} \sqrt{\dfrac{1-\sqrt{1-\dfrac{4E^2 \pi^2 n_1^2 l_q^2 }{t^2 l_p^2}}}{2} + \dfrac{1-\sqrt{1-\dfrac{4E^2 \pi^2 n_2^2 l_q^2 }{t^2 l_p^2}}}{2} + \dfrac{1-\sqrt{1-\dfrac{4E^2 \pi^2 n_3^2 l_q^2 }{t^2 l_p^2}}}{2} + \dfrac{1-\sqrt{1-\dfrac{4E^2 \pi^2 n_4^2 l_q^2 }{t^2 l_p^2}}}{2}} $$
It's quite ugly but it's an equation for $t$.
$$ \dfrac{E^2 \pi^2}{l_p^2} = 2 - \sqrt{\dfrac{1}{4}-\dfrac{E^2 \pi^2 n_1^2 l_q^2 }{t^2 l_p^2}} - \sqrt{\dfrac{1}{4}-\dfrac{E^2 \pi^2 n_2^2 l_q^2 }{t^2 l_p^2}} - \sqrt{\dfrac{1}{4}-\dfrac{E^2 \pi^2 n_3^2 l_q^2 }{t^2 l_p^2}} - \sqrt{\dfrac{1}{4}-\dfrac{E^2 \pi^2 n_4^2 l_q^2 }{t^2 l_p^2}}$$
I don't see how to proceed from here though.
