Let $p$ have roots $r, r, s, t$. Using Vieta's on the coefficient of the cubic and linear terms, we see that $2 r+s+t=r^{2} s+r^{2} t+2 r s t$. Rearranging gives $2 r(1-s t)=\left(r^{2}-1\right)(s+t)$. If $r^{2}-1=0$, then since $r \neq 0$, we require that $1-s t=0$ for the equation to hold. Conversely, if $1-s t=0$, then since $s t=1, s+t=0$ cannot hold for real $s, t$, we require that $r^{2}-1=0$ for the equation to hold. So one valid case is where both these values are zero, so $r^{2}=s t=1$. If $r=\tan y$ (here we stipulate that $0 \leq y<\pi$ ), then either $y=\frac{\pi}{4}$ or $y=\frac{3 \pi}{4}$. In either case, the value of $\tan 2 y$ is undefined. If $r=\tan 2 y$, then we have the possible values $y=\frac{\pi}{8}, \frac{3 \pi}{8}, \frac{5 \pi}{8}, \frac{7 \pi}{8}$. In each of these cases, we must check if $\tan y \tan 3 y=1$. But this is true if $y+3 y=4 y$ is a odd integer multiple of $\frac{\pi}{2}$, which is the case for all such values. If $r=\tan 3 y$, then we must have $\tan y \tan 2 y=1$, so that $3 y$ is an odd integer multiple of $\frac{\pi}{2}$. But then $\tan 3 y$ would be undefined, so none of these values can work. Now, we may assume that $r^{2}-1$ and $1-s t$ are both nonzero. Dividing both sides by $\left(r^{2}-1\right)(1-s t)$ and rearranging yields $0=\frac{2 r}{1-r^{2}}+\frac{s+t}{1-s t}$, the tangent addition formula along with the tangent double angle formula. By setting $r$ to be one of $\tan y, \tan 2 y$, or $\tan 3 y$, we have one of the following: (a) $0=\tan 2 y+\tan 5 y$ (b) $0=\tan 4 y+\tan 4 y$ (c) $0=\tan 6 y+\tan 3 y$. We will find the number of solutions $y$ in the interval $[0, \pi)$. Case 1 yields six multiples of $\frac{\pi}{7}$. Case 2 yields $\tan 4 y=0$, which we can readily check has no solutions. Case 3 yields eight multiples of $\frac{\pi}{9}$. In total, we have $4+6+8=18$ possible values of $y$.