Find the domain of definitions of the functions
(Read the symbols [*] and {*} as greatestintegers and fractional part functions respectively.)
`f(x) = log_(x) sin x`

Answer

To find the domain of the function \( f(x) = \log_x(\sin x) \), we need to consider the conditions under which the logarithm is defined.

### Step-by-Step Solution:

1. **Condition for the Base of the Logarithm**:
The base \( x \) of the logarithm must satisfy two conditions:
- \( x > 0 \)
- \( x \neq 1 \)

2. **Condition for the Argument of the Logarithm**:
The argument \( \sin x \) must also satisfy:
- \( \sin x > 0 \)

3. **Finding the intervals where \( \sin x > 0 \)**:
The sine function is positive in the intervals:
- \( (2k\pi, (2k+1)\pi) \) for any integer \( k \)

4. **Combining the Conditions**:
We need to find the intersection of the intervals where \( \sin x > 0 \) with the conditions for \( x \):
- \( x > 0 \)
- \( x \neq 1 \)

Therefore, we analyze the intervals:
- For \( k = 0 \), the interval is \( (0, \pi) \)
- For \( k = 1 \), the interval is \( (2\pi, 3\pi) \)
- For \( k = 2 \), the interval is \( (4\pi, 5\pi) \)
- And so on...

In each of these intervals, we need to exclude \( x = 1 \) from the interval \( (0, \pi) \).

5. **Final Domain**:
The final domain of the function \( f(x) = \log_x(\sin x) \) is:
- \( (0, 1) \cup (1, \pi) \cup (2\pi, 3\pi) \cup (4\pi, 5\pi) \cup \ldots \)

### Conclusion:
Thus, the domain of the function \( f(x) = \log_x(\sin x) \) is:
\[ \bigcup_{k=0}^{\infty} \left( (2k\pi, (2k+1)\pi) \setminus \{1\} \right) \]