Complete set of solutions in four variables and two nonlinear equations

Question

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?

Answer 1

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 $$ y = (-1)^{n+1}x. $$