(Note: Emil Jann Fiedler found the formula for the Bring quintic using $R(q)$ in 2021, and these two formulas using $\vartheta_3(q)$ and $\vartheta_4(q)$ in 2022.) Recall the Jacobi theta functions,
$$\vartheta_2^4(q) + \vartheta_4^4(q) = \vartheta_3^4(q)$$
and define the usual,
$$\zeta = e^{2\pi i/5},\quad q = e^{\pi i\tau},\quad \tau =\frac{K'(k)}{K(k)}\sqrt{-1}$$
with complete elliptic integral of the first kind $K(k)$. To solve the Bring, one must determine the elliptic modulus $k$. While the two formulas below have a common form, it turns out there may be different equations for $k$. (But why?)
I. Formula using $\vartheta_3(z)$
Given,
$$x^5+5x+c=0$$
then its five roots $x_n$ for $n=(0,1,2,3,4)$ are,
$$\color{blue}{x_n = -c\times\frac{5u^3-5u^2+u}{10u^2-6u+1}\;}$$
where $u$ is solely in $\vartheta_3(z)$,
$$u=\frac{\vartheta_3(\zeta^n q^{1/5})\,\vartheta_3(q^5)-\vartheta_3^2(q)}{\vartheta_3^2(\zeta^n q^{1/5})-4\vartheta_3^2(q)+5\vartheta_3^2(q^5)}$$
and elliptic modulus $k$,
$$k = \sqrt{\frac12+\frac{c\,\sqrt{-c^2+\sqrt{c^4+256}}}{16\sqrt2}}$$
or any appropriate root of the $\color{blue}{octic}$,
$$256k^8 - 512k^6 + (384 + c^4)k^4 - (128 + c^4)k^2 + 16=0$$
Note: It seems this is an improvement over Hermite's method because Hermite's $x_n$ has a square root (hence sign ambiguity) whereas Findler's only has the factor $c$ (which determines the correct sign).
II. Formula using $\vartheta_4(z)$
Given,
$$x^5\color{red}{-}5x+c=0$$
then its five roots are,
$$\color{blue}{x_n = c\times\frac{5v^3-5v^2+v}{10v^2-6v+1}\;}$$
where $v$ is solely in $\vartheta_4(z)$,
$$v=\frac{\vartheta_4(\zeta^n q^{1/5})\,\vartheta_4(q^5)-\vartheta_4^2(q)}{\vartheta_4^2(\zeta^n q^{1/5})-4\vartheta_4^2(q)+5\vartheta_4^2(q^5)}$$
and elliptic modulus $k$ (in trigonometric and radical form),
$$k =\tan\left(\frac14\arcsin\Big(\frac{16}{c^2}\Big)\right)$$
$$k =\sqrt{\frac{\sqrt{2}\,c-\sqrt{c^2+\sqrt{c^4-256}}}{\sqrt{2}\,c+\sqrt{c^2+\sqrt{c^4-256}}}}$$
or any appropriate root of the $\color{blue}{quartic}$,
$$k^4+\left(\tfrac{c}2\right)^2 k^3+2k^2-\left(\tfrac{c}2\right)^2 k+1 =0$$
This is the same quartic in Hermite's method, and in the solution using the Rogers-Ramanujan continued fraction $R(q)$ in this post.
III. Formula using $\vartheta_2(z)$
Since $\vartheta_2^4 + \vartheta_4^4 = \vartheta_3^4$, then the formulas for $\vartheta_2$ and $\vartheta_4$ are similar. But $\vartheta_2$ uses $\tau' = -\dfrac1{\tau}$ and seems to work only for $n=0$.
IV. Questions
The two formulas for $x_n$ have a common form (apart from the sign of $c$) and the two parameters $(u,v)$ also have a common form (apart from the subscript of $\vartheta_n$). And this is the third time the same quartic in $k$ has appeared using different methods. Does this imply there are other methods that employ the octic in $k$?
As before, Findler gave few details. Since this method seems different from Hermite's approach, anyone has a hypothesis how the expressions for $x_n$ and $(u,v)$ were found?
