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This is the edited version of the original problem, hopefully presented in a clearer manner. (I have also renamed this post with a more befitting title)

Problem: y'(x) = 2\sin\left(\frac{y(x)}{2}\right) subjected to boundary conditions: $\lim\limits_{x\to−\infty}y(x)=0$ and $\lim\limits_{x\to+\infty}y(x)=k$ for some constant $k$

Annoying bits:

After integration, I seem to get $y(x)=4\operatorname{arccot}(c\exp(x))$ which has limits $\lim\limits_{x\to−\infty}y(x)=2\pi$ and $\lim\limits_{x\to+\infty}y(x)=0$.

But then I can't fit the boundary conditions of the problem! (This is driving me insane!) Please help. Thanks in advance!

RESOLVED: I have stupidly left out a minus sign, should be $y(x)=4\operatorname{arccot}(c\exp(-x))$

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    @anon: No worries. I just realized that one of the sources of confusion is that I did a change in variable mid way in my calculations hence the jumbled coefficients.2011-08-17

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Let's look at the differential equation y' = 2\sin(y/2). This is autonomous, so

$ \frac{dy}{2\sin(y/2)}=dx.$

The antiderivative of $1/\sin(\cdot)$ is $\ln \tan( \cdot/2)$ (see W|A), so continuing

$ \ln \tan (y/4)=x+C$

$y=4\arctan(Ae^x)+4\pi n$

with $A=e^C$ and $n\in\mathbb{Z}$. The problem is you gave us the expression $\csc+\cot$ (which @J.M. gave an identity for), but this expression has no relation to your differential equation. Like I said, you had stuff going on behind the scenes but weren't completely forward so you weren't able to get the full help you needed. If you make an error and then give us the result of the error, we don't even know something went wrong!

Now we get $\lim_{x\to-\infty}y=4\pi n$, so we can choose $n=0$, and then $\lim_{x\to+\infty}y=\frac{\pi}{2}$, so $k$ can't be arbitrary. The issue here is that your differential equation is first-order, so a single initial condition suffices to determine the solution. Two conditions overdetermines it.

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    Oh, and there are solutions iff $k=2\pi$2011-08-17