On the tribonacci constant with $\cos(2\pi\,k/11)$, plastic constant with $\cos(2\pi\,k/23)$, and others (This post is indebted to Oscar Lanzi.)
Part I. $\color{blue}{p = 11}$
The equation,
$$\big(4\sin[3t] - \tan[t]\big)^2 = 11$$
seems to have five solutions, given by $t = \frac{2\pi\,k}{11}$ for $k = 1,2,3,4,5$, so it ranges from $t = 0.57\; \text{to}\; 2.85$. However, if you look at its plot, it intersects the $x$-axis at SIX points, one of which is less than $0.5,$

As pointed out by Lanzi in an answer, this extra point in fact has a nice closed form involving the tribonacci constant $T$,
$$t = \arccos\big(T/2\big) = 0.40362481\dots$$
the real root of $T^3 = T^2+T+1$.

Part II. $\color{blue}{p = 23}$
I always wonder if things could be generalized, and Lanzi provided the equation for $p=23,$
$$\big(4\sin[2t] + 4\sin[3t] - 4\sin[5t] + 4\sin[6t] - \tan[t]\big)^2=23$$
Analogously, this seems to have eleven solutions, given by $t = \frac{2\pi\,k}{23}$ for $k = 1,2,\dots 11$, so it ranges from $t = 0.27\; \text{to}\; 3.005$. But it intersects the $x$-axis at TWELVE points, one of which is between $1.6$ and $1.9$,

I suspected this would involve the plastic constant $P$, the real root of $P^3 = P+1$, and using Mathematica's FindRoot command and its integer relations subroutine, its closed-form apparently is,
$$t = \arccos\big(\tfrac{1-P}2\big) = 1.73387721\dots$$

Part III. $\color{blue}{p = 31}$
Question: Anybody can find the equation for $p=31$?
(I assume this will involve the supergolden ratio $S$, the real root of $S^3=S^2+1$, but that's just a hunch.)
 A: For $p=31$ the symmetrized sine-tangent relation is
$-\tan(\theta)+4[\sin(2\theta)+\sin(7\theta)+\sin(10\theta)-\sin(12\theta)-\sin(13\theta)+\sin(14\theta)]=(k|31)\sqrt{31}, \theta=2k\pi/31.$
For $k=1,2,3,...,15$ where the Legendre symbol $(k|31)$ is $\pm1$, this gives
$[-\tan(\theta)+4[\sin(2\theta)+\sin(7\theta)+\sin(10\theta)-\sin(12\theta)-\sin(13\theta)+\sin(14\theta)]]^2=31.$
Using appropriate trigonometric identies to express all variables in terms of $x=2\cos(\theta)$ leads to a 30th-degree polynomial in $x$. Since fifteen roots must be in the form $x=2\cos(2k\pi/31)$, the polynomial factors into two fifteenth-degree polynomials, thus:
$P_{15}(x)Q_{15}(x)=0$
where $P$ has the trigonometric roots and $Q$ has the additional coupled roots. Because $Q$ is not a cubic polynomial as was found with $p=11$ and $p=23$, the supergolden ratio fails to emerge immediately. With the 15th-degree polynomial that is actually obtained, three possibilities seem likely:

*

*The supergolden ratio is obtained because $Q$ is further reducible to give the required cubic factor.


*The supergolden ratio is buried in a quintic factor over $\mathbb Q[\psi]$.


*No extraction of the supergolden ratio is possible after all.
UPDATE: It appears, unfortunately, that possibility (3) is the one that holds.

*

*Tito Piezas used Magma to show that $Q$ has no zeroes expressible in radicals, ruling out (1).


*Inspection of computed results for $x=2\cos(\theta)$ from $+2$ to $-2$ in intervals of $0.001$ reveals 24 real roots in this range, meaning no more than three pairs of complex-conjugate roots (the possibility of roots where $x$ is real but $\theta$ is not was not considered). Possibility (2) would have required at least five such pairs, corresponding to quintic factors containing the complex conjugate roots of the supergolden ratio cubic.
Thus the supergolden ratio is not retrieved using the quadratic Gauss sum and associated sine-tangent relation with $p=31$.
However, the supergolden ratio has been obtained from a similar relation with $p=3$. See Tito's answer.
A: Thanks for Lanzi for posting the equation for $p=31$. However, it turns out the supergolden ratio $S$ appeared in $p=3,$
$$\big(4\sin[t]-\tan[t]\big)^2 = 3^3$$
This equation has the unexpected and expected solutions,
$$\begin{align}
t &= \arccos\Big(\tfrac{(S-1)^2}{2}\Big) = 1.46\dots\\
t &= 2\pi/3 = 2.09\dots
\end{align}$$
where $S$ is the real root of $S^3=S^2+1.$ This is clearly seen in the plot below,

A: (This is an addendum to my answer.) It seems best to gather all the equations for small $p$ together.
I. $p=4n+3$
$$\big(4\sin[t] - \tan[t]\big)^2 = \color{blue}3^3$$
$$\big(4\sin[2t] - \tan[t]\big)^2 = 7$$
$$\big(4\sin[3t] - \tan[t]\big)^2 = \color{blue}{11}$$
$$\,\big(4(\sin[6t] - \sin[12t] - \sin[8t]) - \tan[t]\big)^2=\color{blue}{19}\,$$
$$\big(4(\sin[6t] + \sin[7t] - \sin[8t])\, - \tan[t]\big)^2=19$$
$$\big(4(\sin[2t] + \sin[3t] - \sin[5t] + \sin[6t]) - \tan[t]\big)^2=\color{blue}{23}$$
$$\big(4(\sin[2t]+\sin[7t]+\sin[10t]-\sin[12t]-\sin[13t]+\sin[14t])-\tan[t]\big)^2=31$$
Solutions to these equations are $t = \frac{2\pi\,k}p$ for $k = (1,2,\dots \frac{p-1}2)$. However, some remarks:

*

*The cases $p=3,11,23$ have one extra solution $t$ that depend on the supergolden ratio, the tribonacci constant, and the plastic constant, respectively, which are all cubic roots.

*For $p=7$, it intersects the $x$-axis 6 times, but is symmetric in the range $0$ to $\pi$, so involves just three roots.

*The 1st version of $p=19$ is also symmetric, but has two extra solutions given by,$$t = \frac12\arccos\Big(\frac{u_i}{2}\Big)$$ where the $u_i$ are the two real roots of $$u^4 + u^3 - 3u^2 - 4u - 3=0$$ with discriminant $d = 19^2\cdot23.$

*The non-symmetric version of $p=19$ has extra solutions, but is determined by an equation with an unsolvable Galois group and is of $9$-deg.

*The symmetric and non-symmetric versions of $p=31$ has extra solutions determined by two non-solvable equations, with the latter a $15$-deg.

*The symmetric and non-symmetric versions of $p=43$ has extra solutions determined by two non-solvable equations, with the latter a $21$-deg.

And so on? (I used Magma to determine the Galois group and Mathematica to find $p=43, 47.$) From this trend, it would indeed be remarkable if a sine-tangent equation for prime $p>23$ will have extra solutions that depend on radicals.
