Solving a physics problem I got to the following system of equations:
$$\begin{cases} z\cos(y)=c_1 \\ x^2z\sin(y)= c_2 \\ x = c_3\sin(y) \end{cases}$$
I tried many things (replacing them into each other, dividing member by member, using trigonometric identities) but all of them to no avail.
Is it possible to solve this getting to a closed expression for $x$, $y$ and $z$?
Additional info that may be useful (due to the physical nature of the underlying problem):
- All the constants ($c_1$, $c_2$ and $c_3$) are positive.
- $x$ and $z$ must be positive.
- $y$ must be between $0$ and $\pi/2$.
Using Groebner basis methods, I find that $z$ is a root of the sextic polynomial $$ {c_{{3}}}^{4}{z}^{6}-\left(3\,{c_{{1}}}^{2}{c_{{3}}}^{4}{z}+{c_{{2}}}^{2 }\right){z}^{4}+3\,{c_{{1}}}^{4}{c_{{3}}}^{4}{z}^{2}-{c_{{1}}}^{6}{c_{{3}}}^{ 4} $$ $x$ can then be found as $$ x = {\frac {{c_{{3}}}^{3}{z}^{5}}{{c_{{1}}}^{4}c_{{2}}}}+{\frac { \left( - 2\,{c_{{1}}}^{2}{c_{{3}}}^{4}-{c_{{2}}}^{2} \right) {z}^{3}}{{c_{{1}}} ^{4}c_{{2}}c_{{3}}}}+{\frac { \left( {c_{{1}}}^{4}{c_{{3}}}^{4}-{c_{{1 }}}^{2}{c_{{2}}}^{2} \right) z}{{c_{{1}}}^{4}c_{{2}}c_{{3}}}} $$ and $y$ obtained from $\sin(y) = x/c_3$ and $\cos(y) = c_1/z$.
As the equation for $z$ is a cubic in $z^2$, it has closed-form solutions in radicals, but they are not pretty. The discriminant of that cubic is $-c_1^6 c_2^4 c_3^4 (27 c_1^2 c_3^4 + 4 c_2^2)$ which is negative if the $c_i$ are positive, indicating that it has one real solution; since the constant term is negative, that solution is positive.