Complete set of solutions in four variables and two nonlinear equations Consider the system of equations in the four variables $\{x, y, \phi_x, \phi_y \}$:
\begin{align}
x \cos(\phi_x) + y \sin(\phi_y) &= 0 \\
x \sin(\phi_x) - y \cos(\phi_y) &= 0
\, .
\end{align}
It's easy to see that if we set $\phi_x=0$ then the two nontrivial solutions are
\begin{align}
  \phi_y &= \pi/2, \quad y = -x \\
\text{and} \quad
  \phi_y &= -\pi/2, \quad y = x
  \, .
\end{align}
Due to the geometric structure of the problem, I suspect that the general solution is
\begin{align}
  \phi_y &= \phi_x + \pi/2, \quad y = -x \\
\text{and} \quad
  \phi_y &= \phi_x -\pi/2, \quad y = x \tag{$\star$}
\, .
\end{align}
I also noticed that if we square the two equations and add them we get
$$
x^2 + y^2 + 2 x y\left(\cos\phi_x \sin\phi_y - \sin\phi_x \cos\phi_y \right) = 0
$$
which admits the same two solution sets as $(\star)$.
How can we prove that $(\star)$ is or is not the complete set of solutions?
 A: Note that you can rewrite your system as
$$ \begin{bmatrix} \cos(\phi_x) & \sin(\phi_y) \\  -\sin(\phi_x) & \cos(\phi_y) \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = 0. $$
Notice that
$$ \det\begin{bmatrix} \cos(\phi_x) & \sin(\phi_y) \\  -\sin(\phi_x) & \cos(\phi_y) \end{bmatrix} = \cos(\phi_x - \phi_y). $$
We consider two cases:
(1) The matrix is invertible, i.e. $\cos(\phi_x - \phi_y) \not= 0$, or equivalently
$$ \phi_x - \phi_y \not= \frac{\pi}{2} + \pi n \qquad (*) $$
for an integer $n$. In this case we automatically get $x=y=0$, while $\phi_x, \phi_y$ are arbitrary but satisfy $(*)$.
(2) The matrix is not invertible, so
$$ \phi_x = \phi_y + \frac{\pi}{2} + \pi n $$
for an integer $n$. In this case we see that
\begin{align*}
\cos(\phi_x) &= (-1)^{n+1} \sin(\phi_y),\\
\sin(\phi_x) &= (-1)^n \cos(\phi_y),
\end{align*}
so that the system reduces to
\begin{align*}
\sin(\phi_y) ( (-1)^{n+1} x + y ) = 0\\
\cos(\phi_y) ( (-1)^{n+1} x + y ) = 0.
\end{align*}
Both equations are satisfied if and only if
$$ y = (-1)^{n+1}x. $$
