To solve the equation \( \log_{5}(\tan \theta) = \log_{5}(4) \cdot \log_{4}(3 \sin \theta) \) in the interval \( [0, 8\pi] \), we will follow these steps:
### Step 1: Simplify the Equation
Using the change of base formula for logarithms, we can rewrite the equation:
\[
\log_{5}(\tan \theta) = \frac{\log(4)}{\log(5)} \cdot \frac{\log(3 \sin \theta)}{\log(4)}
\]
This simplifies to:
\[
\log_{5}(\tan \theta) = \frac{\log(3 \sin \theta)}{\log(5)}
\]
Thus, we can equate:
\[
\log_{5}(\tan \theta) = \log_{5}(3 \sin \theta)
\]
### Step 2: Remove the Logarithm
Since the logarithm function is one-to-one, we can drop the logarithm:
\[
\tan \theta = 3 \sin \theta
\]
### Step 3: Rewrite the Equation
We can rewrite \( \tan \theta \) in terms of sine and cosine:
\[
\frac{\sin \theta}{\cos \theta} = 3 \sin \theta
\]
This leads to:
\[
\sin \theta = 3 \sin \theta \cos \theta
\]
### Step 4: Rearranging the Equation
Rearranging gives:
\[
\sin \theta (1 - 3 \cos \theta) = 0
\]
This implies two cases:
1. \( \sin \theta = 0 \)
2. \( 1 - 3 \cos \theta = 0 \)
### Step 5: Solve Each Case
**Case 1:** \( \sin \theta = 0 \)
- The solutions are \( \theta = n\pi \) for \( n \in \mathbb{Z} \).
- In the interval \( [0, 8\pi] \), the solutions are \( \theta = 0, \pi, 2\pi, 3\pi, 4\pi, 5\pi, 6\pi, 7\pi, 8\pi \) (total of 9 solutions).
**Case 2:** \( 1 - 3 \cos \theta = 0 \)
- Solving gives \( \cos \theta = \frac{1}{3} \).
- The general solutions for \( \theta \) are:
\[
\theta = \cos^{-1}\left(\frac{1}{3}\right) + 2k\pi \quad \text{and} \quad \theta = -\cos^{-1}\left(\frac{1}{3}\right) + 2k\pi
\]
- In the interval \( [0, 8\pi] \), we find the specific values:
- For \( k = 0 \): \( \theta = \cos^{-1}\left(\frac{1}{3}\right) \) and \( \theta = 2\pi - \cos^{-1}\left(\frac{1}{3}\right) \)
- For \( k = 1 \): \( \theta = \cos^{-1}\left(\frac{1}{3}\right) + 2\pi \) and \( \theta = 2\pi - \cos^{-1}\left(\frac{1}{3}\right) + 2\pi \)
- For \( k = 2 \): \( \theta = \cos^{-1}\left(\frac{1}{3}\right) + 4\pi \) and \( \theta = 2\pi - \cos^{-1}\left(\frac{1}{3}\right) + 4\pi \)
- For \( k = 3 \): \( \theta = \cos^{-1}\left(\frac{1}{3}\right) + 6\pi \) and \( \theta = 2\pi - \cos^{-1}\left(\frac{1}{3}\right) + 6\pi \)
Thus, there are 8 additional solutions from this case.
### Step 6: Total Solutions
Adding the solutions from both cases:
- From \( \sin \theta = 0 \): 9 solutions
- From \( \cos \theta = \frac{1}{3} \): 8 solutions
Total number of solutions in the interval \( [0, 8\pi] \):
\[
9 + 8 = 17
\]
### Final Answer
The number of solutions of the equation in the interval \( [0, 8\pi] \) is **17**.
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