To solve the equation \( \sin(\pi \log_3(1/x)) = 0 \) in the interval \( (0, 2\pi) \), we can follow these steps:
### Step 1: Set up the equation
We start with the equation:
\[
\sin(\pi \log_3(1/x)) = 0
\]
### Step 2: Identify when sine is zero
The sine function is zero when its argument is an integer multiple of \( \pi \):
\[
\pi \log_3(1/x) = n\pi \quad \text{for } n \in \mathbb{Z}
\]
### Step 3: Simplify the equation
Dividing both sides by \( \pi \):
\[
\log_3(1/x) = n
\]
### Step 4: Rewrite the logarithmic equation
Using the property of logarithms, we can rewrite this as:
\[
1/x = 3^n
\]
Thus, we have:
\[
x = 3^{-n}
\]
### Step 5: Determine the range of \( n \)
We need \( x \) to be in the interval \( (0, 2\pi) \):
\[
0 < 3^{-n} < 2\pi
\]
This implies:
\[
3^{-n} < 2\pi \quad \text{and since } 3^{-n} \text{ is always positive, } n \text{ can take negative values.}
\]
### Step 6: Find the bounds for \( n \)
Taking the logarithm base 3 of both sides:
\[
-n < \log_3(2\pi)
\]
Thus:
\[
n > -\log_3(2\pi)
\]
### Step 7: Identify integer values for \( n \)
The integer values of \( n \) can be:
- For \( n = -1 \): \( x = 3^1 = 3 \) (valid since \( 3 < 2\pi \))
- For \( n = 0 \): \( x = 3^0 = 1 \) (valid since \( 1 < 2\pi \))
- For \( n = 1 \): \( x = 3^{-1} = \frac{1}{3} \) (valid since \( \frac{1}{3} < 2\pi \))
- For \( n = 2 \): \( x = 3^{-2} = \frac{1}{9} \) (valid since \( \frac{1}{9} < 2\pi \))
- For \( n = 3 \): \( x = 3^{-3} = \frac{1}{27} \) (valid since \( \frac{1}{27} < 2\pi \))
- Continuing this, all negative \( n \) will yield valid \( x \) values.
### Step 8: Calculate the sum of the roots
The valid roots are:
\[
x = 3^{-n} \quad \text{for } n = -1, 0, 1, 2, \ldots
\]
Thus, the roots are:
\[
3^1, 3^0, 3^{-1}, 3^{-2}, \ldots = 3, 1, \frac{1}{3}, \frac{1}{9}, \ldots
\]
### Step 9: Sum the roots
The sum of the roots can be expressed as:
\[
S = 3 + 1 + \left( \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots \right)
\]
The series \( \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots \) is a geometric series with first term \( a = \frac{1}{3} \) and common ratio \( r = \frac{1}{3} \).
The sum of an infinite geometric series is given by:
\[
S = \frac{a}{1 - r} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}
\]
Thus, the total sum of the roots is:
\[
S = 3 + 1 + \frac{1}{2} = 4 + \frac{1}{2} = \frac{8}{2} + \frac{1}{2} = \frac{9}{2}
\]
### Final Answer
The sum of all the roots of the equation in the interval \( (0, 2\pi) \) is:
\[
\frac{9}{2}
\]
