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.


Numerator is not a problem, since it is the sum of two positive quantities $${\log}^{2}(2\cos x -1) + e^{x}>0, \forall x\in\mathbb{R}$$

except the values which give negative or null $2\cos x -1$

Notice that $\sqrt{1-\sin^2 x} - \sin(x)=\cos x -\sin x$

Argument of logarithm, denominator and exponent lead to the system

$ \left\{ \begin{array}{l} 2\cos x -1>0\\ \cos x - \sin x >0 \\ 0\le\sqrt{x}\le 1\\ \end{array} \right. $

$ \left\{ \begin{array}{l} 0\le x \le 1\\ \cos x >\dfrac{1}{2}\\ \dfrac{\cos x}{\cos x} - \dfrac{\sin x}{\cos x} >0 \end{array} \right. $

$ \left\{ \begin{array}{l} 0\le x \le 1\\ \cos x >\dfrac{1}{2}\to -\dfrac{\pi}{3}+2k\pi< x <\dfrac{\pi}{3}+2k\pi,\,k\in\mathbb{Z}\\ \tan x < 1 \to h\pi\le x <\dfrac{\pi}{4}+h\pi,\,h\in\mathbb{Z} \end{array} \right. $

which gives the conditions

$\color{red}{0\leq x<\dfrac{\pi}{4}}$


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