Csdrhrt

So as the question says finding domain of-

$f(x) = log(log_{|sin x|}(x^2-8x+23)-largefrac{3}{log_{2}|sin x|})$

$large f(x)=log(log_{|sin x|}(x^2-8x+23)-largefrac{3}{log_{2}|sin x|})$

What I did-

$largeimplies log_{|sin x|} (x^2 - 8x + 23) - frac{3}{ (log_{2} |sin x|)} > 0$

$large implies log_{|sin x|} (x^2 - 8x + 23) - 3 log_{|sin x|} 2 > 0$

How to proceed further?

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:
$$
left{begin{matrix}
|sin x| > 0\
|sin x| neq 1\
end{matrix}right. iff left{begin{matrix}
sin x neq 0\
sin x neq pm 1\
end{matrix}right. implies x neq kfrac{pi}{2}, k in mathbb{Z}$$

Passing to the arguments:
$$left{begin{matrix}
x^2 - 8x + 23 > 0\
|sin x| > 0
end{matrix}right. iff left{begin{matrix}
forall x in mathbb{R}\
x neq kpi, k in mathbb{Z}
end{matrix}right. implies x neq kpi, k in mathbb{Z}$$
This is clearly a subset of the previous solution, so until now we have
$$x neq kfrac{pi}{2}, k in mathbb{Z} tag{1}$$

It remains to apply the constraint to the first $ln$:
$$log_{|sin x|} (x^2 - 8x + 23) - frac{3}{log_{2} |sin x|} > 0\
frac{ln(x^2-8 x+23)-3ln2}{ln|sin x|}>0$$

Since $ln|sin x|$ is always negative (except when $|sin x| = 1$, which we already excluded before) we can write:
$$ln(x^2-8 x+23)-3ln2<0\
ln(x^2-8 x+23)<ln8\
x^2-8 x+15<0\
3 < x < 5$$
Putting this last result together with $(1)$ we get:
$$3 < x < 5 land x notin left{pi, frac{3pi}{2}right}$$

I plotted it with Desmos and got this:
https://www.desmos.com/calculator/ld5vmseh9y

$$
begin{align}
log{log_{|sin x|}(x^2−8x+23)−frac{3}{log_2|sin x|}} &=
log{log_{|sin x|}(x^2−8x+23)+log_{|sin x|}frac{1}{8}} \
&= log{log_{|sin x|}frac{(x^2−8x+23)}{8}} \
&=log{frac{logfrac{(x^2−8x+23)}{8}}{log|sin x|}}
end{align}
$$
$$
=>frac{logfrac{(x^2−8x+23)}{8}}{log|sin x|} > 0\
text{In the above equation the denominator, } log{|sin x|} text{, is always negative.}\ text{ So the numerator must also be negative, }
logfrac{(x^2−8x+23)}{8} < 0 \
begin{align}
&=> frac{x^2−8x+23}{8} < 1 \
&=> {x^2−8x+23} < 8 \
&=> {x^2−8x+15} < 0 \
&=> (x-3)(x-5)<0 \
end{align} \
x in(3,5) \
text{Also we have a constraint on x from the initial equation : } |sin x|>0,ne 1
\ => x in (0,2pi) - {frac{pi}{2},pi,frac{3pi}{2}}
\ x in(3,5) - {pi,frac{3pi}{2}}
$$

$|sin x|>0\implies xne npi$ where $ninmathbb{I}$

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$log_2|sin x|ne0\ implies |sin x|ne1\implies xne (2n+1)large{fracpi2}$
where $ninmathbb{I}$

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$x^2-8x+23>0\implies xinmathbb{R}$

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$log_{|sin x|}(x^2-8x+23)-largefrac{3}{log_{2}|sin x|}>0 \implies
log_{|sin x|}(x^2-8x+23)-largelog_{|sin x|}8>0\implies
large{log_{|sin x|}(frac{x^2-8x+23}8)>0}\implies
frac{x^2-8x+23}8<1\implies
x^2-8x+23<8\implies
x^2-8x+15<0\implies
xin(3,5)$

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Combining all $$LARGE xin(3,5)-{pi,frac32pi}$$