Solution to $\frac{x}{\tan\left(\frac{\pi}{2}-\frac{\pi}{x}\right)}=\pi$ In the following equation, is it possible to solve for numerical value of $x$.
$$\frac{x}{\tan\left(\frac{\pi}{2}-\frac{\pi}{x}\right)}=\pi$$
 A: Because $\tan\left(\dfrac{\pi}{2}-\alpha\right)=\operatorname{cotan}(\alpha)=\dfrac{1}{\tan(\alpha)}$, the given equation is equivalent to:
$$\tan\left(\dfrac{\pi}{x}\right)=\dfrac{\pi}{x}$$
Setting $a=\dfrac{\pi}{x}$, we have to look for non-zero solutions $a_k \ (k \in \mathbb{Z^*})$ of equation  $\tan(a)=a$. There are an infinity of them: see graphics below. 
As a consequence, the solutions of the initial problem are $x_k=\dfrac{\pi}{a_k}.$
Let us fix $k$; a very efficient way to obtain $a_k$ is to invert relationship 
$$\tag{1}\tan(a)=a.$$ 
One might be tempted to invert (1) by writing plainly into $a=\tan^{-1}(a)$.
In fact, due to the periodicity of tangent function, inversion of (1) is 
$$\tag{2}a=f_k(a) \ \ \  \text{where} \ \ \ f_k(a):=\tan^{-1}(a)+k \pi$$ 
(as can be understood by the graphical representation of $f_k$, which is the $k$th branch in blue on the graphics, resulting from a $k$ times "upper translation" from mother function $\tan^{-1}$). 
Solving (2) is easy by using the classical "fixed point" sequence built in this way:
$$\alpha_{p+1}=f_k(\alpha_{p}) \ \ \ p=0,1,2...$$
(with any initial $\alpha_0$). It converges very rapidly to the solution $a=a_k$ of (2) due to a small value of the derivative of $f_k$ in the vicinity of this root $a_k$.
Remarks: 
1) Two very interesting references : this (contains very interesting pointers) and this, the latter with a solution using Lagrange inversion formula.
2) I have used notation $\tan^{-1}$ because it is now very widespread (maybe due to hand calculators ?), although notation $\operatorname{atan}$ or $\operatorname{arctan}$ is less misleading . More precisely, $y=\tan(x)$ has $f_k(x)$ as its reciprocal function for its $k$th "branch". Sometimes one meets the concept of a unique  multivalued inverse function ; the natural context in which this concept has its full meaning is complex function theory.

A: One cannot solve this, but with some simple rewriting, we have
$$x=\pi\tan\left(90-\frac{180}x\right)$$
Since $\tan(\theta)$ for any $\theta\in\mathbb Q$, and where we use degrees, we have
$$\tan(\theta)\in\mathbb A$$
That is, it will be an algebraic number.
Now, that means that since $\pi$ is transcendental, $\pi\tan\left(90-\frac{180}x\right)$ must be transcendental.
Thus, $x$ cannot be any rational number.  It is also not a multiple of $\pi$, since we are in radians.  So the answer is not trivial.
It is also not possible to solve, due to the fact that if $\theta$ is not rational, $\tan(\theta)$ has no closed form, working in radians.  For example, noone knows what $\tan(\sqrt2)$ is exactly, so if $x$ is not a rational number... then we cannot solve the equation in closed form.
Looking at a graph, we have an infinite amount of interception points:

Indeed, the behavior of the $\tan$ around $x=0$ is not very good for most numerical methods.
A: I started trying out different values for $x$ and seeing how much effect they have on the problem
So after fiddling with that for a wile I found that as the value of $x$ gets closer and closer to infinity. the equation actually gets closer and closer to being true (that is closer to equaling $\pi$). so if $x$ is equal to infinity then the equation must be true (or equal to $\pi$). So $x$ in this equation must be infinity. But that also explains why the equation is impossible to solve because you cant work out the tangent for. $$\left(\frac{\pi}{2}-\frac{\pi}{\infty}\right)$$
