How to solve $x+\sin(x)=b$ How can I solve $x+\sin(x)=b$ for $x \in [0,π]$?  
We take $b \in [0,π]$. I don't know how to find the solution.
 A: You can't actually solve it, but you can approximate.
Here is the expansion for $\require{cancel}\sin(x)$ around $x=\pi/2$:
$$\sin(x)=1-\frac{(x-\frac\pi2)^2}{1\times2}+\frac{(x-\frac\pi2)^4}{1\times2\times3\times4}-\dots$$
To give a crude estimate,
$$\sin(x)\approx1-\frac{(x-\frac\pi2)^2}2$$
So let us solve the approximate problem:
$$x+\left(1-\frac{(x-\frac\pi2)^2}2\right)=b$$
$$\frac12x^2-\left(1+\frac\pi2\right)x+\frac{\pi^2}4-1+b=0\tag{expand}$$
$$x\approx1+\frac\pi2\pm\sqrt{3+\pi-2b}\tag{Quadratic Formula}$$
Given some guessing, you could determine if the $+\sqrt{}$ or the $-\sqrt{}$ is the one you are looking for.
Use longer expansions of sine for better approximations.
For example, if $b=1$, then
$$x\approx1+\frac\pi2\pm\sqrt{1+\pi}=\begin{cases}0.5357\\\cancel{4.6059}\end{cases}$$
See that $x\cancel\approx4.6059$, and so the other root is the correct approximation.  Plugging it in gives $1.0462$, a decent approximation to start with.

You could also use numerical methods like Newton's method, as mentioned above in the comments.
Applying Newton's method gives the following algorithm:
$$x_{n+1}=x_n-\frac{x_n+\sin(x_n)-b}{1+\cos(x_n)}$$
For $b=1$ and initial guess being $x_0=0.5357$, we get
$$\begin{align}
x_0 & =0.5357 \\
x_1 & =0.5109 \\
x_2 & =0.5110 \\
x_3 & =0.510973429\dots
\end{align}$$
Indeed, taking that final value for $x$ gives $x_3+\sin(x_3)=1$ for as many digits my calculator takes.
