how can I solve x = t + sin(t) for t? I am trying to understand how to approach solving the equation in the title. Any suggestions how to do it?
solve $x = t + \sin(t)$ for $t$
an approximation method that I can script into Matlab would be fine
 A: Assuming you are trying to solve for $t$ as a function of $x$:
We cannot solve for $t$ using only standard functions. We can, though, approximate what $t$ is, particularly when $x$ is large, by using the fact that $-1\leq \sin t \leq 1$.
Then, $$x = t + \sin(t)\iff t = x - \underbrace{\sin t}_{\in [-1,1]} \implies x - 1 \leq t \leq x+1$$
For VERY large $x$, the impact of $\sin t$ on the value $x - \sin t$ will be very minimal.

Given your edit (since rolled back): 
If you are hoping to determine the range of $x$, given $\,t \in [0, \pi/2]$, e.g., then $0 \leq \sin t \leq 1$. 
So the range of $x$ would be from $x = 0 + \sin 0 = 0\;$ to $\;x = \pi/2 + \sin(\pi/2) = \dfrac{\pi}2 + 1 = \dfrac{\pi + 2}{2}$.
Hence, $$x \in \left[0, \dfrac{\pi + 2}{2}\right]$$  Without knowing anything more about $t$, that's about all we can say.
A: You cannot express it in terms of sum of finite number of elementary functions. 
A: You can approximate the value of the expression around a given point, $a$.
$t+\sin(t) \approx t+(t-a)+\sin(a)+\cos(a)\cdot(t-a)- \sin(a)\cdot (t-a)^2 \frac{1}{2!}-\cos(a)\cdot (t-a)^3\frac{1}{3!}.... $ 
This is called the Taylor approximation. If $a=0$ then it's usual to named Maclaurin approximation.
A: *

*For transcendental equations in general, numerical approximation methods such as Newton's (as well as others: see link) are usually employed. 

*For small values of t, use the Taylor series for $\sin t\simeq t-\dfrac{t^3}{3!}$ , and then either apply the cubic formula yourself, or even ask MatLab to solve it directly, since it possesses in-build numerical methods for solving general n-th degree polynomials. 

*For greater values of the argument, we have $t\simeq x$.
