Find the domain of the function,
`f(x)=log{log_(abs(sinx)){x^(2)-8x+23)-(3)/(log_(2)abs(sinx))}`.

Answer

To find the domain of the function

\[ f(x) = \log\left(\frac{\log_{|\sin x|}(x^2 - 8x + 23) - 3}{\log_2 |\sin x|}\right), \]

we need to ensure that all the components of the function are defined and valid.

### Step 1: Conditions for the logarithm

1. **Base of the logarithm**: The base \( |\sin x| \) must be positive and not equal to 1. This gives us two conditions:
- \( |\sin x| > 0 \) which implies \( \sin x \neq 0 \).
- \( |\sin x| \neq 1 \) which implies \( \sin x \neq \pm 1 \).

From \( \sin x \neq 0 \), we have:
\[ x \neq n\pi \quad \text{for } n \in \mathbb{Z}. \]
From \( \sin x \neq \pm 1 \), we have:
\[ x \neq \frac{(2n+1)\pi}{2} \quad \text{for } n \in \mathbb{Z}. \]

### Step 2: Argument of the logarithm

2. **Argument of the logarithm**: The argument of the logarithm must be greater than zero:
\[ \frac{\log_{|\sin x|}(x^2 - 8x + 23) - 3}{\log_2 |\sin x|} > 0. \]
This implies:
\[ \log_{|\sin x|}(x^2 - 8x + 23) - 3 > 0 \quad \text{and} \quad \log_2 |\sin x| > 0. \]
From \( \log_2 |\sin x| > 0 \), we have:
\[ |\sin x| > 1 \quad \text{(which is impossible)}. \]
Therefore, we only need to consider \( \log_{|\sin x|}(x^2 - 8x + 23) > 3 \).

### Step 3: Solve the inequality

3. **Solving the inequality**:
\[ \log_{|\sin x|}(x^2 - 8x + 23) > 3 \implies x^2 - 8x + 23 > |\sin x|^3. \]
The quadratic \( x^2 - 8x + 23 \) is always positive since its discriminant is negative:
\[ D = (-8)^2 - 4 \cdot 1 \cdot 23 = 64 - 92 = -28 < 0. \]
Thus, \( x^2 - 8x + 23 > 0 \) for all \( x \).

### Step 4: Final domain

4. **Combining conditions**: The conditions we have are:
- \( x \neq n\pi \) for \( n \in \mathbb{Z} \).
- \( x \neq \frac{(2n+1)\pi}{2} \) for \( n \in \mathbb{Z} \).

Therefore, the domain of the function can be expressed as:
\[ \text{Domain } f(x) = \mathbb{R} \setminus \{ n\pi, \frac{(2n+1)\pi}{2} \mid n \in \mathbb{Z} \}. \]