I'm trying to find the domain of the following function:
$$f(x)={\Biggl(\frac{\displaystyle\ {\log}^{2}(2\cos(x)-1) + e^{x}}{\displaystyle \sqrt{1-\sin^2(x)} - \sin(x)}\Biggr)}^{\arcsin\sqrt{x}}$$
I have reasoned this way: since I have an exponential function with variable base, this last one must be posed $>0$. And so:
$$\frac{\displaystyle\ {\log}^{2}(2\cos(x)-1) + e^{x}}{\displaystyle \sqrt{1-\sin^2(x)} - \sin(x)}>0$$
Even the exponent has an its condition which is: $$0\le\sqrt{x}\le1$$
Hence I have to solve the following system:
$$ \left\{ \begin{array}{c} \frac{\displaystyle\ {\log}^{2}(2\cos(x)-1) + e^{x}}{\displaystyle \sqrt{1-\sin^2(x)} - \sin(x)}>0 \\ 0\le\sqrt{x}\le1 \end{array} \right. $$
This is the point in which I get blocked, I do not know how to solve the first inequality of the system: specifically, I'm having troubles in solving the numerator: $${\log}^{2}(2\cos(x)-1) + e^{x}>0$$ I tried to solve it by using graphic mode but I found it not so accurate.
Now, someone could say me if my calculus is wrong or could give me another way (or hint) to determine the domain of $f(x)$? Thank you.