This is what I did: Let $\frac{dx}{dt} = \frac{1}{1+\frac{1}{2}\cos x}$, then $\frac{du}{dt} = 0$. Separation of variables for the first item gives $x+\frac{1}{2}\sin x = t + c$, so $u$ is constant along this curve. With some initial data we are done.
However it is a homework question, saying
Show that the solution is given by $u(x, t) = u_0(\xi)$ where $\xi$ is the unique solution of $\xi+\frac{1}{2}\sin \xi = x+\frac{1}{2}\sin x -t$.
I don't understand. Shouldn't the solution be $u(x,t) = u_0(x+\frac{1}{2}\sin x -t)$?
Not exactly. The general solution is, better, $u(x,t) = f(x+\frac{1}{2}\sin x -t)$ for some $f$, being $u_0(x)=u(x,0)$ the given initial data. So,
$u_0(x)=f(x+\frac{1}{2}\sin x)$. The same is $u_0(\xi)=f(\xi+\frac{1}{2}\sin \xi)$. This equality is exactly what determines the form $f$ has with these initial data:
let $X=\xi+\frac{1}{2}\sin \xi$, then, $u_0(\xi)=f(X)$. We can see more easily $\xi$ as the unique solution of $X=\xi+\frac{1}{2}\sin \xi$.
Now, remember that we have in general that $u(x,t) = f(x+\frac{1}{2}\sin x -t)$, so is,
$u(x,t)=f(X)=u_0(\xi)$ with $X=x+\frac{1}{2}\sin x -t$ or, equivalently, with $\xi$ as the unique solution of $\xi+\frac{1}{2}\sin \xi = x+\frac{1}{2}\sin x -t$.
You can keep track of the steps by giving some cool name to the inverse of the function $g(x)=x+\frac{1}{2}\sin x$