BG71 wrote: The real numbers $a$, $b$ and $c$ verify that the polynomial $p(x)=x^4+ax^3+bx^2+ax+c$ has exactly three distinct real roots; these roots are equal to $\tan y$, $\tan 2y$ and $\tan 3y$, for some real number $y$. Find all possible values of $y$, $0\leq y < \pi$. Let $r_1=\tan y,r_2=\tan 2y,r_3=\tan 3y,r_4=\tan u$ be the four roots of the quartic (where $u\in\{y,2y,3y\}$) The presence of same coefficient $a$ means equation $r_1+r_2+r_3+r_4=r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4$ 1) If $r_1r_2+r_2r_3+r_3r_1=1$ and so $r_1+r_2+r_3=r_1r_2r_3$ $r_1+r_2=0$ would imply $r_1r_2=1$, impossible and so $r_1+r_2\ne 0$ $r_1r_2=1$ would imply $r_3(r_1+r_2)=0$ from the first equation and so $r_3=0$ and $r_1+r_2=0$ from the second equation. Impossibe And so both $r_3=\frac{1-r_1r_2}{r_1+r_2}$ and $r_3=\frac{r_1+r_2}{r_1r_2-1}$ and so $r_3^2=-1$ (multiplying), impossible 2) If $r_1r_2+r_2r_3+r_3r_1\ne 1$ So $r_4=\frac{r_1+r_2+r_3-r_1r_2r_3}{r_1r_2+r_2r_3+r_3r_1-1}$ I let you prove the identity : $\frac{\tan (u-v)+\tan(u)+\tan(u+v)-\tan(u-v)\tan(u)\tan(u+v)}{\tan(u-v)\tan(u)+\tan(u)\tan(u+v)+\tan(u+v)\tan(u)-1}=-\tan(3u)$ Applying this to $u=2y$ and $v=y$, this becomes $r_4=-tan(6y)$ And so the three families of solutions : $\tan y=\tan(-6y)$ and $y=\frac{k\pi}7$ with $k\in\{1,2,3,4,5,6\}$ $\tan 2y=\tan(-6y)$ and $y=\frac{k\pi}8$ with $k\in\{1,3,5,7\}$ (restricted to odd values in order to avoid undefined tangeants). $\tan 3y=\tan(-6y)$ and $y=\frac{k\pi}9$ with $k\in\{1,2,3,4,5,6,7,8\}$ Hence the solutions $\boxed{\frac{\pi}7,\frac{2\pi}7,\frac{3\pi}7,\frac{4\pi}7,\frac{5\pi}7,\frac{6\pi}7,\frac{\pi}8,\frac{3\pi}8,\frac{5\pi}8,\frac{7\pi}8,\frac{\pi}9,\frac{2\pi}9,\frac{3\pi}9,\frac{4\pi}9,\frac{5\pi}9,\frac{6\pi}9,\frac{7pi}9,\frac{8\pi}9}$

Minor typo in last line (second to last number) $\frac {7 \pi}{9}$