Using a straight edge and a pair of compasses only, it is imporssible to trisect angles like $\pi /3$. In fact, if we want the $x$ in $4x^3-3x-c=0,c=\cos 3\alpha$ to be constructible, we must have at least one rational root. Otherwise the degree of field extension $\mathbb Q(x)$ will be $3$, which is not a power of $2$, hence not constructible.
Question: how can we find the set of all angles that can be trisected?
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Remember that we do not require the original or trisected angle to be constructible on their own. For example, $\frac\pi7$ is trisectible since the difference between this angle and the constructible $\frac\pi6$ is $\frac\pi{42}$, which can be doubled to give the trisected angle $\frac\pi{21}$.
Wikipedia gives a characterisation of all trisectible angles as those angles $\theta$ where $4x^3-3x-\cos\theta$ is reducible over $\mathbb Q(\cos\theta)$. This differs from the characterisation of all angles whose trisections are constructible by adjoining $\cos\theta$ to $\mathbb Q$.
Yes. See Wikipedia on constructible angles You can construct any angle that is $2\pi$ divided by a Fermat prime, $2^{2^n}+1.$ We know of $3,5,17,65537$, but there may be others. Then you can bisect and add these as much as you want.