So as the question says finding domain of-
$f(x) = \log(\log_{|\sin x|}(x^2-8x+23)-\large\frac{3}{\log_{2}|\sin x|})$
$\large f(x)=\log(\log_{|\sin x|}(x^2-8x+23)-\large\frac{3}{\log_{2}|\sin x|})$
What I did-
$\large\implies \log_{|\sin x|} (x^2 – 8x + 23) – \frac{3}{ (\log_{2} |\sin x|)} > 0$
$\large \implies \log_{|\sin x|} (x^2 – 8x + 23) – 3 \log_{|\sin x|} 2 > 0$
How to proceed further?
Best Answer
To determine the domain of $log_a(b)$, you need to apply the constraints: $$b > 0\\a>0 \land a \neq 1$$.
So let's apply them to the bases first: $$ \left\{\begin{matrix} |\sin x| > 0\\ |\sin x| \neq 1\\ \end{matrix}\right. \iff \left\{\begin{matrix} \sin x \neq 0\\ \sin x \neq \pm 1\\ \end{matrix}\right. \implies x \neq k\frac{\pi}{2}, k \in \mathbb{Z}$$
Passing to the arguments: $$\left\{\begin{matrix} x^2 - 8x + 23 > 0\\ |\sin x| > 0 \end{matrix}\right. \iff \left\{\begin{matrix} \forall x \in \mathbb{R}\\ x \neq k\pi, k \in \mathbb{Z} \end{matrix}\right. \implies x \neq k\pi, k \in \mathbb{Z}$$ This is clearly a subset of the previous solution, so until now we have $$x \neq k\frac{\pi}{2}, k \in \mathbb{Z} \tag{1}$$
It remains to apply the constraint to the first $\ln$: $$\log_{|\sin x|} (x^2 - 8x + 23) - \frac{3}{\log_{2} |\sin x|} > 0\\ \frac{\ln(x^2-8 x+23)-3\ln2}{\ln|\sin x|}>0$$
Since $\ln|\sin x|$ is always negative (except when $|\sin x| = 1$, which we already excluded before) we can write: $$\ln(x^2-8 x+23)-3\ln2<0\\ \ln(x^2-8 x+23)<\ln8\\ x^2-8 x+15<0\\ 3 < x < 5$$ Putting this last result together with $(1)$ we get: $$3 < x < 5 \land x \notin \left\{\pi, \frac{3\pi}{2}\right\}$$
I plotted it with Desmos and got this: https://www.desmos.com/calculator/ld5vmseh9y