Can this equation be solved? $x+\sin(x)=\frac{11\pi}{48}$ So I was revisiting an older problem and seeing if I could solve it in a different way. I boiled the equation down to this: $$x+\sin(x)=\frac{11\pi}{48}$$ I can't imagine how to isolate x, and a number of computer solvers also broke down in the effort. Desmos can solve it, but does not give the precise value. Is there a way to do it? Thanks.
 A: $f(x)=x+\sin x$ is monotonically increasing, and without bounds, so there will be exactly one solution.
$y=\frac{11\pi}{48}$ is (relatively) small, so a first guess is obtained by setting $\sin x\approx x$ to get $x\approx \frac{11\pi}{96}$. An improved value is obtained by including the next term of the sine series,
$$
2x-\frac16x^3\approx y\implies x\approx \frac y2 + \frac{y^3}{96}.
$$
This gives the numerical value $x=0.3638613210103829$ which is already close to the (more) exact value $0.363965532996313$.

Instead of including more terms of the sine series, one could also directly iterate the fixed point equation 
$$
x=g(x)=\frac12(y+x-\sin(x))
$$
A: Hint.-The equation being trascendental we make numerical calculation and stop where we consider enough approximation. For this there are several ways. Here one.
First $a=\dfrac{11\pi}{48}\approx0.7194831$. Second in the neighborhood of $0$ one has $x\approx\sin(x)$ so  $x+x=2x\approx0.7194831\Rightarrow x\approx0.359974$ which is a first approximation.
Now for $f(x)=x+\sin(x)$ we have successive values
$$f(0.36)\approx0.712...\lt a\\f(0.365)\approx0.72...\gt a\\f(0.364)\approx0.72...\gt a\\f(0.363)\approx0.718...\lt a$$ Trying with $x$ from $0.3635$ till $0.3639$ we have $f(x)\lt a$ then with $x=0,36395$ till $0.363965$ we get
$$f(0.363965)\approx 0.719947\approx a$$ 
We stop at this approximation $x\approx0.363965$
