prove that $x(t) \in ]0,\pi[$ Given the Cauchy Problem:
$
\left\{
\begin{array}{@{}l}
x'(t) = \sin (x(t)),\ t\in\mathbb{R}\\ 
x(0)=x_0 \in ]0,\pi[
\end{array}
\right.
$
I try to prove that $x(t) \in ]0,\pi[ $
What I did:
Using Cauchy-Lipschitz theorem, I proved that the Cauchy Problem has a unique solution on $\mathbb{R}$
I also proved that the solution cannot be constant since no $k \in \mathbb{Z}$ verifies $x_0=k\pi$.
I don't know to follow-up from there. any help much appreciated.
 A: It is an easy consequence of uniqueness of solutions. The constant functions $0$ and $\pi$ are solutions of the equation. By uniqueness, the graph of the solution with $x(0)\in(0,\pi)$ cannot cross the graph of those constant solutions (otherwise there would be two constant solutions through the same point.) This implies that $0<x(t)<\pi$.
A: We have $$\frac{dx}{dt}=\sin x,$$ so $$\frac{dx}{\sin x}=dt$$ and the equation is easily solvable in the form $t=\varphi(x).$ You can now find a domain of $\varphi$ taking into account the initial condition.
A: Below is the explicit solution. Probably that will help you investigate the values of $x(t)$. 
$$x'(t)=\sin x(t)\Rightarrow x''(t)=\cos x(t)\cdot x'(t)=\cos x(t)\sin x(t)$$
On the other hand
$$x'(t)^2=\sin x(t) x'(t)\Rightarrow \int x'(t)^2\,dt=\int \sin x(t) x'(t)\,dt$$
But
$$\int \sin x(t) x'(t)\,dt=\int\sin x(t)\,dx(t)=-\cos x(t)+C$$
and after several steps of integration by parts we get
$$\int x'(t)^2\,dt=\int x'(t)\,d x(t)=x'(t)x(t)-\int x(t)x''(t)\,dt\\
=x'(t)x(t)-\int \cos x(t) x(t)x'(t)\,dt=x'(t)x(t)-\int\cos x(t)x(t)\,d x(t)\\
=x'(t)x(t)-\frac{1}{2}x^2(t)\cos x(t)-\frac{1}{2}\int x^2(t)\sin x(t)\,dt\\=x'(t)x(t)-\frac{1}{2}x^2(t)\cos x(t)-\frac{1}{2}\int x^2(t)\,dx(t)\\
=x(t)\sin x(t)-\frac{1}{2}x^2(t)\cos x(t)-\frac{1}{6}x^3(t)$$
Overall we get
$$x(t)\sin x(t)-\frac{1}{2}x^2(t)\cos x(t)-\frac{1}{6}x^3(t)=-\cos x(t)+C$$
where $C$ is uniquely determined by $x(0)=x_0$ and equals
$$C=x_0\sin x_0-\frac{1}{2}x^2_0\cos x_0-\frac{1}{6}x^3_0+\cos x_0$$
