To solve the equation \( \sin(2\theta) = 1 \), we follow these steps:

1. **Identify the general solution for \( \sin(x) = 1 \):**
   The sine function equals 1 at specific angles. Specifically, \( \sin(x) = 1 \) when:
   \[
   x = \frac{\pi}{2} + 2k\pi
   \]
   where \( k \) is any integer.

2. **Apply this to the equation \( \sin(2\theta) = 1 \):**
   Set \( 2\theta = \frac{\pi}{2} + 2k\pi \).

3. **Solve for \( \theta \):**
   Divide both sides by 2:
   \[
   \theta = \frac{\pi}{4} + k\pi
   \]
   where \( k \) is any integer.

Thus, the general solution for \( \theta \) is:
\[
\theta = \frac{\pi}{4} + k\pi
\]
where \( k \) is any integer.

Would you like more details on this solution or have any other questions?

Here are some related questions:
1. What is the general solution for \( \sin(x) = -1 \)?
2. How would the solution change if the equation was \( \sin(2\theta) = -1 \)?
3. What is the general solution for \( \cos(2\theta) = 1 \)?
4. How do you solve \( \tan(2\theta) = 1 \)?
5. What is the graphical interpretation of \( \sin(2\theta) = 1 \)?

**Tip:** Always consider the periodic nature of trigonometric functions when solving equations involving sine, cosine, or tangent.

Explore the step-by-step solution for the trigonometric equation sin(2θ) = 1, covering key trigonometric concepts such as the sine function, general solutions of trigonometric equations, and the periodicity of the sine function. Suitable for students in Grades 10-12.

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