For the second equation, sin(𝜃) = 1, the solution in [0, 2𝜋) is as follows.

To solve the equation \(\sin(\theta) = 1\) in the interval \([0, 2\pi)\), we follow these steps:

1. **Understanding the equation**: The sine function, \(\sin(\theta)\), represents the y-coordinate of a point on the unit circle. We are looking for the angle(s) \(\theta\) where the sine value equals 1.

2. **Identifying the solution**: The sine function reaches its maximum value of 1 at \(\theta = \frac{\pi}{2}\). This is the only point in the interval \([0, 2\pi)\) where \(\sin(\theta) = 1\).

3. **Conclusion**: Therefore, the solution to the equation \(\sin(\theta) = 1\) in the interval \([0, 2\pi)\) is:
   \[
   \theta = \frac{\pi}{2}
   \]

Would you like more details or have any questions?

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Here are 8 related questions you might find useful:

1. What are the general solutions for \(\sin(\theta) = 1\) on the real number line?
2. How do you solve \(\sin(\theta) = -1\) in the interval \([0, 2\pi)\)?
3. What are the solutions to \(\sin(\theta) = 0\) in \([0, 2\pi)\)?
4. How do the graphs of \(\sin(\theta)\) and \(\cos(\theta)\) compare in the interval \([0, 2\pi)\)?
5. What is the general solution to \(\sin(\theta) = c\) for any constant \(c\)?
6. How would you solve the equation \(\sin(2\theta) = 1\) in \([0, 2\pi)\)?
7. What is the derivative of the sine function, and how does it relate to solving \(\sin(\theta) = 1\)?
8. How does the unit circle help in understanding the solutions of trigonometric equations?

**Tip:** Remember, the sine function is periodic with a period of \(2\pi\), meaning it repeats every \(2\pi\) units. This property is key when solving trigonometric equations over different intervals.

Learn how to solve the trigonometric equation sin(theta) = 1 within the interval [0, 2π) using detailed step-by-step instructions. Understand the properties of the sine function and its behavior on the unit circle. Suitable for high school students studying trigonometry.

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