Is it possible to solve $\cos(x) + 2e^{x} = 0$ analytically?

Question

My Calculus textbook uses $f(x) = \sin(x) + e^{2x}$ as an example of a function with infinitely many local extrema. That much is clear, because $\cos(x) + 2e^{x} =0 $ has infinitely many solutions for $x<0$. I'm curious if there is a way to solve \begin{equation} \cos(x) + 2e^{x} =0 \end{equation} exactly. The only thought I've had so far is to write this as \begin{equation} e^{ix} + e^{-ix} + 4e^{x} = 0 \end{equation} and go from there, but I haven't had much luck.

Answer 1

$$f(x_n)=\cos(x_n)+2e^{x_n}$$ $$f'(x_n)=2e^{x_n}-\sin(x_n)$$ now use the iterative formula: $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ $$x_{n+1}=x_n-\frac{2e^{x_n}+\cos(x_n)}{2e^{x_n}-\sin(x_n)}$$ this is probably the easiest numerical way to solve it

Edit:

In terms of using: $$e^{ix}+e^{-ix}+4e^x=0$$ we can express this as: $$\left(e^x\right)^{i}+\left(e^x\right)^{-i}+4e^x=0$$ so if we let $u=e^x$ we get: $$u^i+u^{-i}+4u=0$$ which I do not think can be solved

Answer 2

Here's the closest I can think of to a solution without a computer. Note that on $[0,\infty)$, $e^x$ outgrows $\cos x$ quickly. It is easy to verify, thus that $2e^x+\cos x$ is always greater than $0$ in this range (just check a few derivatives). If $x$ is negative with a large magnitude, then $e^{x}$ is tiny, so the solution to your equation is akin to the solutions of $\cos x=0$, which as you know are given by $x=\pi(n+1/2)$ for integral $n$. Meanwhile, if $x$ is negative but with a small magnitude, perhaps you can try linear approximations for the function near the roots of $\cos x$, and see how much that is expected to change due to the added part of $e^x$.