Find the domain of the function, inverse trigonometric function 
So I am asked to find the domain of this function. For the bottom part, I know the domain is $x\neq \frac{\pi}{4}+n\pi$ but I don't really know how to deal with the $\arcsin(\frac{x}{2}+3)$ and then combining the domains after. Any guidance? Thanks.
 A: For the numerator,
$$-1\le\frac{x}{2}+3\le1$$
$$-4\le\frac{x}{2}\le{-2}$$
$$-8\le x\le-4$$
For the denominator, we must have $$\ln(\sin(2x))\ne0$$
which gives
$$\sin(2x)\ne1$$
which gives $$x\ne\frac{\pi}4+n\pi$$
where $n\in \mathbb Z$.
However, we must also have $$\sin(2x)\gt0$$
as the domain of $\ln(x)$ is $x\gt0$.
So, 
$$0\lt\sin(2x)\lt1$$
$$2x\in(2n\pi,(2n+1)\pi)-\frac{(4n+1)\pi}{2}$$
$$x\in(n\pi,(n+\frac12)\pi)-(n+\frac14)\pi$$
or
$$x\in(n\pi,(n+\frac14)\pi)\cup((n+\frac14)\pi,(n+\frac12)\pi)$$
where $n\in \mathbb Z$
The solution is then the intersection of the domains corresponding to the numerator and the denominator.
A: Domaine of numerator: $$-1\le \frac x2+3\le 1\iff -8\le x\le -4$$
Domaine of denominator (where the sinus is positive): $$\bigcup_{k\in\Bbb Z}\space]k\pi,k\pi+\frac{\pi}{2}[$$
Hence the domaine $D$ of the functions is
$$D=[-8,\frac{-5\pi}{2}[\space\cup\space ]-2\pi,\frac{-3\pi}{2}[\space\setminus\{\frac{-7\pi}{4}\}$$ (it is excluded the point $\frac{-7\pi}{4}$ because $\sin(\frac{-7\pi}{2})=1$ giving denominator zero).
A: Note that the denominator is only defined when $\boxed{\sin2x>0}$ and $\boxed{\sin2x\ne1}$, and you just considered the latter inequality.
The arcsine is only defined when its argument is in the interval $[-1,1]$, so you must have
$$
-1\le\frac{x}{2}+3\le 1
$$
(it's $x$, not $\pi$). This is the same as $-8\le x\le -4$. Combine with the condition from the denominator and you're done.
A: Hint:
The argument of $\arcsin(X)$ must be such that $|X|\le 1$,  because the $\sin$ function have allways values in $[-1,1]$ .
So the conditions such that $f(x)$ is a real function are expressed by the system:
$$
\begin{cases}
\left | \frac{x}{2}+3\right|\le 1\\
0>\sin 2x <1
\end{cases}
$$
where the second inequality expresses the condition that the $log$ function is defined as a real number and it is not null.
Can you solve this system? ( Note that your condition for the denominator is wrong because does not exclude the case $<0$).
