Calculate the domain of Irrational function, which contains Log and Trigonometric functions I'm trying to find the domain of the following function:
$$f(x)=\Biggl( {\frac{\displaystyle\log(\arctan(x))}{\displaystyle \sqrt{\sin(x)-\frac{1}{2}} + \sin(x)}}\Biggr)^{1/3}  $$
I reasoned this way: 
$$
\left\{ 
\begin{array}{c}
\arctan(x)>0 \\ 
\sqrt{\sin(x)-\frac{1}{2}}+\sin(x)\neq0 \\ 
\sin(x)\ge\frac{1}{2}
\end{array}
\right. 
$$
and so:
$$
\left\{ 
\begin{array}{c}
x>0 \\ 
x\neq2kπ+\arcsin(\frac{1+\sqrt3}{2})\\
x\neq2kπ+π-\arcsin(\frac{1+\sqrt3}{2})\\
x\neq2kπ+\arcsin(\frac{1-\sqrt3}{2})\\
x\neq2kπ+π-\arcsin(\frac{1-\sqrt3}{2}) \\ 
2kπ + \frac{π}{6}\le x\le\frac{5}{6}π+2kπ
\end{array}
\right. 
$$
And here I get blocked. Consulting Wolfram Alpha, my result seems to be wrong.
Someone could say me where my calculus are wrong or could give me another way to determine the domain of $f(x)$? Thank you.

Since $\sqrt{\sin x-\frac{1}{2}}\ge0$, if $\sin x\ge\frac{1}{2}$, then it is guaranteed that $\sqrt{\sin x-\frac{1}{2}}+\sin x\ge\frac{1}{2}$, and therefore is nonzero. Thus, we only need $x>0$ and $\sin x\ge\frac{1}{2}$; the other inequality is redundant. This gives us $$\left\{x:2k\pi+\frac{\pi}{6}\le x\le2k\pi+\frac{5\pi}{6} \ \text{for some integer} \ k\ge0\right\}.$$
