<s>[INST] How to solve $\cos({r^{-4}}\cos 4\theta - {4\theta}) = 0$ for $\theta$ and $r$? Problem Statement

How does one solve the trigonometric equation
  $$\cos({r^{-4}}\cos 4\theta - {4\theta}) = 0$$ for $\theta$ and $r$?

My Try
Taking the inverse cosines of both sides, I get
$${r^{-4}}\cos 4\theta - {4\theta} = \cos^{-1}(0) = \frac{\pi}{2}+k\pi.$$
Rearranging the last equation gives
$$\frac{\cos 4\theta}{4 \theta} - r^4 = \frac{{r^4}\left(\frac{\pi}{2}+k\pi\right)}{4 \theta}.$$
This is where I get stuck.
QUERY

Of course, I know that I can rewrite the last equation as
  $$\cos 4\theta = {r^4}\left(4\theta + \frac{\pi}{2} + k\pi\right).$$

Unfortunately, I do not know what to do or what approach to take past this point.  Should I use iteration to approximate the solution(s), if any?  Or is it (logically) possible to prove that there do not exist any solutions to this equation?


Thanks!
Added September 15 2017
Actually, I have
$$r = (x^2 + y^2)^{1/2} := f(x, y)$$
and
$$\theta = \tan^{-1}\left(\frac{y}{x}\right) := g(x,y).$$
Does that help?
 [/INST] Notice that, from the equation
$$\cos 4\theta = {r^4}\left(4\theta + \frac{\pi}{2} + k\pi\right),$$
and the equation
$$r = (x^2 + y^2)^{1/2}$$
I get that $r \geq 0$, so that
$$\cos 4\theta \geq 0.$$
This last inequality implies that
$$-\frac{\pi}{2} \leq 4\theta \leq \frac{\pi}{2}$$
which implies that
$$-\frac{\pi}{8} \leq \theta \leq \frac{\pi}{8}.$$
(Note that, when $r = 0$, then $\theta = -\pi/8, \pi/8$.)
We now bound $r$ from above, given that
$$r^4 = \frac{\cos 4\theta}{4\theta + \frac{\pi}{2} + k\pi}.$$
An upper bound for $\cos 4\theta$ is given by $1$.  However, there is no lower bound for $4\theta + \frac{\pi}{2} + k\pi$, since $k \in \mathbb{Z}$.
Thus, there is no global maximum for
$$\frac{\cos 4\theta}{4\theta + \frac{\pi}{2} + k\pi}.$$
Hence, $r^4$ is not bounded from above.
Consequently, we conclude that the equation
$$\cos 4\theta = {r^4}\left(4\theta + \frac{\pi}{2} + k\pi\right)$$
is solvable when
$$\frac{-\pi}{8} \leq \theta \leq \frac{\pi}{8}.$$
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