To find the domain of the function \( f(x) = \frac{1}{1 - \sin(2x)} \), we need to determine the values of \( x \) for which the denominator is non-zero.

### Step-by-step:

1. **Denominator condition**:
   \[
   1 - \sin(2x) \neq 0
   \]
   This implies:
   \[
   \sin(2x) \neq 1
   \]
2. **Solving for \( x \)**:
   \[
   \sin(2x) = 1 \quad \text{when} \quad 2x = \frac{\pi}{2} + 2n\pi, \quad n \in \mathbb{Z}
   \]
   Therefore:
   \[
   x = \frac{\pi}{4} + n\pi, \quad n \in \mathbb{Z}
   \]

### Conclusion:
The domain of \( f(x) \) is all real numbers except where:
\[
x = \frac{\pi}{4} + n\pi, \quad n \in \mathbb{Z}
\]

In set notation:
\[
\text{Domain of } f(x): \mathbb{R} \setminus \left\{ \frac{\pi}{4} + n\pi \mid n \in \mathbb{Z} \right\}
\]

Would you like a deeper explanation or examples for this?

Here are 5 related questions:
1. What is the range of \( f(x) = \frac{1}{1 - \sin(2x)} \)?
2. How do you find the domain of more complex trigonometric functions?
3. What happens if the denominator contains \( \cos(2x) \) instead of \( \sin(2x) \)?
4. How do transformations affect the domain of trigonometric functions?
5. What is the period of \( f(x) \) in this case?

**Tip**: When finding domains of fractions, always look for values that make the denominator zero.

Discover how to find the domain of the trigonometric function f(x) = 1 / (1 - sin(2x)). This solution breaks down the step-by-step process to determine which x-values make the denominator non-zero. Suitable for students in Grades 10-12.

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