I am Only in Algebra II Trig and am just getting into these kinds of things. So please forgive me if my terminology or notation are wrong.
So, Say I have $$\sum_{n=0}^\infty \frac{(-1)^{n} x^{2n+1}}{(2n+1)!}$$ Which is the Taylor series for $\sin x$
But I take that and set it equal to $$\sum_{n=1}^\infty \frac{1}{2^n}$$
Which obviously would evaluate to 1, (evaluate is the word?) anyways, how would I solve for $x$ when it's inside of a series (series is the word?) Basically I'm evaluating $\sin x = 1$ using the series for $\sin$
The standard way of "solving for an $x$ inside a sum" would be to find a closed form for that sum, and then solve an equation. As you probably have guessed, there are cases when this is not possible. In such cases, one can come up with approximate solutions looking some special or dominant terms of the series.
If you want to solve $\sin x=1$, using series seems overkill, and actually it seems the other way around. Just solve this as a regular equation. This equation can be solved looking at the behaviour of the sine function, which is a periodic function. You know by trigonometry that $\sin(\pi/2)=1$, so $\pi/2$ is a solution. Do you know of other values that evaluate to 1? Think of the unit circle.