Help finding the solution of the following equation system. I have the following equation with some trigonometric functions:
( where: $u_{x}, u_{y}, a, b$ are known. To be found are: $\theta,\phi$)
$$
\begin{cases} u_{x} = a\sin(\theta)\cos(\phi) + b\sin(\phi) \\ u_{y} = b\sin(\theta)\cos(\phi) - a\sin(\phi) \end{cases} 
$$
It seems at first simple but i cannot find the solution.
What I did is the following:


*

*I squared both equations and I got:
$$
\begin{cases} u_{x}^{2} = ( a\sin(\theta)\cos(\phi) + b\sin(\phi))^{2} \\ u_{y}^{2} = ( b\sin(\theta)\cos(\phi) - a\sin(\phi) )^{2} \end{cases} 
$$

*adding both equations toghether and then simpifing:
$$
u_{x}^{2} + u_{y}^{2} = a^{2}\sin(\theta)^{2}\cos(\phi)^{2} + b^{2}\sin(\phi)^{2} + b^{2}\sin(\theta)^{2}\cos(\phi)^{2} + a^{2}\sin(\phi)^{2}
$$

*I get the following equation:


$$
\frac{u_{x}^{2} + u_{y}^{2}}{a^{2} + b^{2}} = \sin(\theta)^{2}\cos(\phi)^{2} + \sin(\phi)^{2}
$$
but I cannot get the final solution.
Could you please help me to solve this problem?
Regards and thanks

rewrite like follows
$u_x-b\sin(\phi)=a\sin(\theta)\cos(\phi)$ (1)
$u_y+a\sin(\phi)=b\sin(\theta)\cos(\phi)$ (2)
and divide (1) by (2)
