I want(ed) to analytically solve the transcendental equation $e^{\sin(x)} = \sin(e^x)$ for a closed form solution.
My working so far is:
$\frac{e^{ie^x}-e^{-ie^x}}{2i}=e^{\frac{e^{ix}-e^{-ix}}{2i}}$
$\frac{e^{2ie^x}-1}{2ie^{ie^x}}=e^{\frac{e^{2ix}-1}{2ie^{ix}}}$
$e^{2ie^x}-1 =2i(e^{ie^x+\frac{e^{2ix}-1}{2ie^{ix}}})$
$e^{2ie^x} = 2ie^{\frac{e^{2ix}-2e^{x + ix}-1}{2ie^{ix}}}+1$
$2ie^x = \ln(2ie^{\frac{e^{2ix}-2e^{x + ix}-1}{2ie^{ix}}}+1)$
I might've gotten a little bit further if I carried on but I couldn't find any summation-inside-logarithm theorems so I gave up - and I assume this is an unsolvable transcendental equation.
I know that most transcendental equations are unsolvable - barring lucky exceptions where terms can be canceled. And when I post these equations asking for questions on Quora, StackExchange etc. they immediately recognize the equation has no elementary solutions by just looking at it and post numerical approximations instead - is there a series of tests I can carry out on an equation to see if the transcendental equation has any elementary solutions? I'm not actually looking for approximations or practical takeaways, I'm just trying to practice my algebra skills for college.