How would you solve an equation of the form $e^{-x} - \sin(x) = 0$? I've been trying to do this for ages. I'm worried that it's impossible, but I have heard that it can be done by hand. 
As long as I can get $x$ by hand I can obviously work out its value via calculator.
Please help, I am so stuck and desperately need assistance.
Thanks in advance!
 A: First of all, that equation is highly nonlinear, as a result analytically we may not (can not) find a solution.
First Observation: Note that when $x$ becomes very large $e^{-x}$ becomes very small so $$e^{-x}-\sin x\approx0-\sin x$$ and hence the zeros of $\sin x$ are the zeros of $e^{-x}-\sin x$. This observation also ensures that solution for $e^{-x}-\sin x=0$ exists.
Second Observation: For $x<0$ $e^{-x}$ is an increasing function and dominates $\sin x$ (Which can be seen in the graph too) so there are no zeros of the function $e^{-x}-\sin x$ for $x<0$.
So what's the solution? Well, the solution can only be found by Numerical methods. One easy way to do that by plotting graphs.

This is the graph of $e^{-x}-\sin x$ where the first observation is visualized.

This is the graph showing the intersection of $e^{-x}$ [Blue line] and $\sin x$ [Red line].
In one word the solutions are $x=0.589,3.096,6.285$ etc.
Hope this works.
A: You cannot find a closed form for the zeroes, but you can find asymptotic approximations.
Let $x_n$ be the zero in $(\pi n-\frac \pi 2, \pi n+\frac \pi 2)$. Write $x_n=\pi n+\varepsilon_n$ with $|\varepsilon_n|<\frac \pi 2$.
Then $$e^{-\pi n -\varepsilon_n}=\sin(\pi n+\varepsilon_n)=(-1)^n\sin(\varepsilon_n)\tag{1}$$
Taking the limit, the left-hand-side converges to $0$. So this implies that $\varepsilon_n\rightarrow 0$. Thus, using Taylor expansions in $(1)$,
$$e^{-\pi n}(1+o(1))= (-1)^n \varepsilon_n(1 +o(1))$$
which means that
$$\varepsilon_n=(-1)^ne^{-\pi n}(1+o(1))$$
$$\boxed{x_n=\pi n+(-1)^n e^{-\pi n} + o(e^{-\pi n})}$$
The zeroes therefore converge exponentially fast towards multiples of $\pi$.
You can keep playing that game and plug the residual back into $(1)$ to find the next term in the expression of $x_n$, but this is already converging quite fast.

