Domain of $\sqrt{\log_x(\cos2\pi x)}$ What is the domain of $\sqrt{\log_x(\cos2\pi x)}?$
This is how i solved it.
$\log_x(\cos2\pi x)\geq 0\Rightarrow \frac{\log(\cos2\pi x)}{\log x}\geq 0$
$\frac{\log(\cos2\pi x).\log x}{\log^2 x}\geq 0$
$\log(\cos2\pi x).\log x\geq 0$
which gives me no common solution.But the answer given in the book is $(0,\frac{1}{4})\cup (\frac{3}{4},1)\cup \left\{x:x\in \mathbb N,x\geq 2\right\}$.
Please help me.
 A: The numerator $\log (\cos (2\pi x))>0$ if and only if $\cos (2\pi x) \geq 1$, or $\cos (2\pi x)=1$. Hence $x \in \mathbb{N}$. The denominator $\log x>0$ if and only if $x>1$. However we must  exclude the points where $\cos (2\pi x)\leq 0$. The domain is therefore
$$
\{2,3,4,\ldots\} \cup (0,1/4) \cup ( 3/4,1)
$$
A: Your work is correct up until you deduce there are no solutions. Note that in order to divide by $\log x$, we know $\log x ≠ 0$ so $x≠ 1$. From $(\log x)(\log \cos 2π x) \geq 0$ we have multiple cases to consider.


*

*$$\log x > 0 \iff x>1$$ This tells us $\log (\cos 2π x) \geq 0$, i.e. $\cos 2π x \geq 1$. Thus $x ∈ \{ … -2,-1,0,1,2,…\}$, but also $x>1$ giving $x ∈ \{ 2,3,4,…\}$.

*$$\log x < 0 \iff 0<x<1$$ This means $\log(\cos 2π x) \leq 0$, i.e. $0 < \cos 2π x \leq 1 $. Drawing the quadrants we see that $2π x$ must lie in the first or fourth quadrant, i.e. $x∈ (0,1/4) ∪ (3/4,1)$.


Since any of these $x$'s work, the maximum domain of $\sqrt{\log_x \cos 2π x}$ is the union of these,
$$ \{ 2,3,4,… \} ∪ (0,1/4) ∪ (3/4,1)$$
