I have been trying to solve this equation that involves exponentiation and trigonometry.
$$k e^{k\theta+2k\pi} \cos \theta - e^{k\theta+2k\pi} \sin \theta + \sin \theta = 0$$
Where $k$ is a non-integer constant and $\theta$ is a variable. I need to solve this for $\theta$.
This equation is for $dx/d\theta$ in a modified logarithmic spiral and I try to determine when the spiral tangent line is vertical.
I have tried a few things but have been unable to solve it.
A @Ivan Kaznacheyeu commented, using $\theta=t-2\pi$, we need to find the zeros of function $$f(t)=e^{k t} (k \cos (t)-\sin (t))+\sin (t)$$ for, I presume, $0 \leq t \leq 2\pi$.
The first thing which is interesting to look at is the zero level curve of $f(t)$ (have loook here). I shall suppose that we are concerned by the root for $t \in (0,\pi)$.
Assuming $k>0$, the first estimate of the solution is $$t_0=\tan ^{-1}(k)$$ The first iterate of Newton method is $$t_{(2)}=\tan ^{-1}(k)+\frac{k}{\left(k^2+1\right) e^{k \tan ^{-1}(k)}-1}$$
Since $$f(t_0) \times f''(t_0)=-k^2 \left(\frac{1}{k^2+1}+2 e^{k \tan ^{-1}(k)}\right)~~< ~~0$$ by Darboux theorem, $t_{(2)}$ is an overestimate of the solution.
This could be improved using instead the first iterate of Halley method
$$t_{(3)}=\tan ^{-1}(k)+\frac{2 k \left(\left(k^2+1\right) e^{k \tan ^{-1}(k)}-1\right)}{k^2+2 \left(k^2+1\right) e^{k \tan ^{-1}(k)} \left(k^2+\left(k^2+1\right) e^{k \tan ^{-1}(k)}-2\right)+2}$$
We could still improve computing $t_{(4)}$, the first interate of Householder method (formula not typed here since bit too long for the page) and even for the next higher orders. All the expressions of $t_{(n)}$ are fully explicit in terms of $k$ and basic functions of it.
This does not look too bad, at least for $k\geq $ as shown below
$$ \left( \begin{array}{ccccccc} k & t_{(2)} & t_{(3)} & t_{(4)} & t_{(5)}& t_{(6)}& \text{solution} \\ 1.00 & 1.0806830 & 0.9924595 & 1.0077207 & 1.0053055 & 1.0056672 & 1.0056210 \\ 1.25 & 1.0784245 & 1.0387710 & 1.0435558 & 1.0430623 & 1.0431095 & 1.0431055 \\ 1.50 & 1.0964786 & 1.0783341 & 1.0798824 & 1.0797763 & 1.0797828 & 1.0797827 \\ 1.75 & 1.1228185 & 1.1145118 & 1.1150178 & 1.1149943 & 1.1149953 & 1.1149953 \\ 2.00 & 1.1518166 & 1.1480411 & 1.1482055 & 1.1482002 & 1.1482004 & 1.1482004 \\ 2.25 & 1.1806697 & 1.1789696 & 1.1790222 & 1.1790211 & 1.1790211 & 1.1790211 \\ 2.50 & 1.2080044 & 1.2072455 & 1.2072621 & 1.2072619 & 1.2072619 & 1.2072619 \\ 2.75 & 1.2332202 & 1.2328839 & 1.2328890 & 1.2328890 & 1.2328890 & 1.2328893 \\ 3.00 & 1.2561380 & 1.2559898 & 1.2559914 & 1.2559913 & 1.2559913 & 1.2559914 \\ 3.25 & 1.2768018 & 1.2767367 & 1.2767372 & 1.2767372 & 1.2767372 & 1.2767372 \\ 3.50 & 1.2953646 & 1.2953361 & 1.2953362 & 1.2953362 & 1.2953362 & 1.2953362 \\ 3.75 & 1.3120244 & 1.3120119 & 1.3120119 & 1.3120119 & 1.3120119 & 1.3120119 \\ 4.00 & 1.3269887 & 1.3269832 & 1.3269832 & 1.3269832 & 1.3269832 & 1.3269832 \\ 4.25 & 1.3404565 & 1.3404541 & 1.3404541 & 1.3404541 & 1.3404541 & 1.3404541 \\ 4.50 & 1.3526098 & 1.3526087 & 1.3526087 & 1.3526087 & 1.3526087 & 1.3526087 \\ 4.75 & 1.3636107 & 1.3636102 & 1.3636102 & 1.3636102 & 1.3636102 & 1.3636102 \\ 5.00 & 1.3736011 & 1.3736009 & 1.3736009 & 1.3736009 & 1.3736009 & 1.3736011 \end{array} \right)$$
Edit
For the fun, we also could write is that the solution is the (impossible) inverse of $$k=\tan(t)+\frac 1 t W\left(-t \,\tan (t)\,e^{-t\, \tan (t)} \right)$$ where $W(.)$ is Lambert function.
But this shows that, for $k>0$ there is no solution smaller than the root of $$t \, \tan(t)=1 $$ that is to say $t=0.8603335890\cdots$