Lindemann's proveproof 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
$$ \frac{lcm(\theta,2\pi)}{\theta} = 3k, k\in\mathbb{N}$$$$ \frac{\operatorname{lcm}(\theta,2\pi)}{\theta} = 3k, k\in\mathbb{N}$$
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, alltogetheraltogether 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:
http://en.wikipedia.org/wiki/Angle_trisectionhttps://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.