Solve $\sin x = 1 - x$

How would you be solve sin x = 1 - x, without drawing the graph and manually measuring the point of intersection?

There's no closed-form analytical solution for this kind of transcendental equation. You can either solve it graphically (as you said) or numerically. There are various methods for solving it numerically, see this Wikipedia article for an overview.

If you're not looking for a particularly efficient solution, the simplest thing to do in this case might be to write the equation as

$$x=1-\sin x$$

and iterate the map

$$x_{n+1}=1-\sin x_n$$

until you achieve satisfactory convergence. This is very inefficient, though; it only makes sense if you do it on a computer and don't mind if it takes a lot of operations; you'll need a lot of patience to get anywhere with it with a calculator. :-)

A far more efficient method that yields $x$ to double-precision machine accuracy within a couple of iterations would be Newton's method:

$$f(x) = x + \sin x - 1 \stackrel{!}{=} 0\;,$$

$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n - \frac{x+\sin x - 1}{1 + \cos x}\;.$$

Below is a calculus-type approach. Remark that to first order $\sin x \simeq x,$ so your result will be somewhere around $x = \frac{1}{2}.$ That suggests a Taylor expansion of sine around $\frac{\pi}{6} \simeq \frac{1}{2}.$ You can introduce $\epsilon = x - \frac{\pi}{6},$ such that $$\sin(x) = \frac{1}{2} + \frac{\sqrt{3}}{2}\epsilon - \frac{1}{4}\epsilon^2 + \mathcal{O}(\epsilon^3).$$ Identifying $1 - x = 1 - (\frac{\pi}{6} + \epsilon),$ you can equate both sides of your equation and solve a linear, quadratic, ..., polynomial equation in $\epsilon$, which will however involve knowing a reasonable approximation to $\pi$ and/or $\frac{\pi}{6}.$