https://file.aiquickdraw.com/mathbot/file/1727727221159rcltfxdt.jpg

The equation in the image is:

\[
\ln(\sin x) = 0
\]

Let's solve this equation step by step:

### Step 1: Exponentiate both sides
Since the natural logarithm (ln) and exponentiation are inverse operations, we can exponentiate both sides of the equation:

\[
e^{\ln(\sin x)} = e^0
\]

This simplifies to:

\[
\sin x = 1
\]

### Step 2: Solve for \(x\)
Now, we need to find the values of \(x\) such that \(\sin x = 1\).

The sine function equals 1 at \(x = \frac{\pi}{2} + 2n\pi\), where \(n\) is any integer. Therefore, the general solution is:

\[
x = \frac{\pi}{2} + 2n\pi \quad \text{for} \quad n \in \mathbb{Z}
\]

Thus, \(x\) can take values like \(\frac{\pi}{2}, \frac{5\pi}{2}, \frac{9\pi}{2}, \dots\).

Would you like more details or have any questions?

Here are 5 related questions to consider:
1. What are other ways to solve trigonometric equations like this one?
2. How does the natural logarithm function behave for negative arguments?
3. Can \(\ln(\sin x) = 0\) have a solution when \(x\) is not within the principal range of sine?
4. How does the sine function behave for angles greater than \(2\pi\)?
5. What is the period of the sine function, and how does it affect the general solution?

**Tip**: When solving trigonometric equations, always consider the periodic nature of the trigonometric functions to find all possible solutions.

This solution demonstrates how to solve the equation ln(sin(x)) = 0. The process involves logarithmic properties and the periodic nature of trigonometric functions, specifically the sine function. Suitable for advanced high school or undergraduate students.

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