Lindemann's proof of the transcendence of $\pi$ has settled the question, whether an arbitrary angle can be trisected, using straightedge and compass alone, to the negative.  
In the following, _trisectable_ always means with straightedge and compass alone.

I could however find nothing but some vague statements about angles, that are _trisectable_; a typical such statement is, that up to a few exceptions, it is impossible to trisect angles.

**Question:** I would therefore like to know, whether the angles, that are _trisectable_ have been fully characterized and, if yes, who has done that first. 

**Edit:**  
in view of the valuable feedback I got, I will try to summarize things from my perspective
  
-   it can be proven, that an angle $\theta$ can be constructed if and only if $\cos(\theta) \in \mathbb{E},$ where $\mathbb E$ (for Euclidean), often called the "constructable numbers," is the smallest subfield of the real numbers that is closed under taking square roots of positive elements.    

-  there seems to be common agreement, that an angle $\theta$ can only be trisected with straightedge and compass, if $\cos(\frac{\theta}{3}))$ is constructable. 

So either being trisectable and constructable are equivalent or, there are angles that are not constructable, but can be trisected.

I suspect however, that there are certain angles (e.g. $\frac{2\pi}{7+\frac{1}{3}}$), that can be trisected despite not being constructable, namely angles $\theta$, for which

$$ \frac{\operatorname{lcm}(\theta,2\pi)}{\theta} = 3k, k\in\mathbb{N}$$  

those angles, which I would like to call _auto-trisecting_, are then either all constructable or, there are exceptions which are counter examples to the characterisation via constructability (the notion of being _auto-trisecting_ of course easily generalizes to being _auto-$n$-secting_).

Interestingly, the only "basic" (i.e. involving only a single Fermat prime) constructable angles, that can be trisected, but are not _auto-trisecting_, seem to be the ones of the form$$3k\frac{2\pi}{2^nF_0}, k\in\mathbb{N},n\in\mathbb{N}_0, F_0 := 2^{2^0}+1=3$$  
With that observation in mind, and assuming that all trisectable angles are either constructable or _auto-trisecting_, a bullet-proof strategy for finding a trisection of an angle, that is known to be trisectable with straightedge and compass alone, is to simply to construct multiples of it, until a period has been completed.  The so generated multiples either 

- subdivide the angle in case it is _auto-trisecting_ and the size of a trisection is equal to the sum of $\frac{1}{3}$ of the (equal) subdivisions or,

- do not subdivide the angle, but constitute to a constructable regular n-gon and, by determining n, it is possible to determine a constructable angle that resembles a trisection. 

in both cases it is possible to find a trisection and, altogether it can be said that the topic of trisecting an angle in case that its trisectability is known, can be discussed at classroom-level.

As a remark it can be said, that there is no hope for proving the existence of further Fermat primes by demonstrating that the known Fermat primes together with _auto-trisecting_ angles do not cover all cases of trisectable angles and thus would necessitate the existence of further Fermat primes.

__Edit II:__  
I found the following Wiki article that elucidates the history of proving that angles are not trisectable in general:  
https://en.wikipedia.org/wiki/Angle_trisection  
according to that article, the proof was found by Pierre Wantzel in 1837 and is not based on Lindemann's proof of the transcendence of $\pi$, but rather on Galois theory.