In searching extremes of Chebyshev-functions of the second kind, I encountered the equation $\frac{\sin((k+1)\theta)}{\sin(k(\theta))}=\frac{k+1}{k}$. For $k=2$ one gets the polynomial equation $2z^2+z+2=0$ and the complex roots $z=-\frac{1}{4}+\frac{i\sqrt{15}}{4}$ and its conjugate. Since the roots have norm $1$, I wondered whether the argument $\theta$ of $z$ in polar form can be expressed as a rational fraction of $\pi$.
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$\begingroup$If $\cos\theta=-\frac{1}{4}$ then $\theta\not\in\pi\mathbb{Q}$. Let $f(x)=2x^2-1$, $a_0=-\frac{1}{4}$ and $a_{n+1}=f(a_n)$. The first terms of this sequence are $$ a_0=-\frac{1}{4},\qquad a_1=-\frac{7}{8},\qquad a_2=\frac{17}{32},\qquad a_3=-\frac{223}{512} $$ and it is simple to check by induction that $\nu_2(a_n)$ is decreasing towards $-\infty$. In particular the sequence $\cos\theta,\cos(2\theta),\cos(4\theta),\cos(8\theta),\ldots$ is not periodic and $\theta$ cannot be a rational multiple of $\pi$.
As an alternative approach, $\cos\theta=-\frac{1}{4}$ and $\theta\in\pi\mathbb{Q}$ imply that $\frac{-1+\sqrt{-15}}{4}$ is a root of unity. $\frac{-1+\sqrt{-15}}{4}$ is algebraic over $\mathbb{Q}$, but it is not an algebraic integer, hence it cannot be a root of unity.
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