$\arctan$ of a square root as a rational multiple of $\pi$

Solution 1:

The answer is NO in general : the converse of your statement does not hold. For example, if $\tan(x)=\sqrt{2}$ then $2\cos(x)=\pm\frac{2}{\sqrt{3}}$ which is algebraic but not an algebraic integer, so $x$ cannot be a rational multiple of $\pi$ (see the first answer to that related question for example).

If $\tan(x)=\sqrt{\frac{p}{q}}$ where $p$ and $q$ are coprime positive integers, you will have $2\cos(x)=\pm \sqrt{\frac{4q}{p+q}}$, and this will be an algebraic integer iff $p+q$ divides $4q$, iff $p+q$ divides $4$ (since $p+q$ is coprime to $q$), iff $p+q$ is one of $1,2$ or $4$, iff $(p,q)$ is one of $(1,1),(1,3)$ or $(3,1)$.

Solution 2:

The complete theory analogous to the one for rational tangents is worked out in

On Angles Whose Squared Trigonometric Functions are Rational
by John H. Conway, Charles Radin, Lorenzo Sadun
http://arxiv.org/abs/math-ph/9812019