Solution:

Answer: 18

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 $2 r(1 - s t) = (r^{2} - 1)(s + t)$.

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 an 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 $(r^{2} - 1)(1 - s t)$ 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 $y$.