Because the sine function repeats its values every 2𝜋 units, we get all solutions of the equation by adding integer multiples of 2𝜋 to the solutions found in the previous step. Thus, if we let k be any integer, we should add 2k𝜋 to each of the three solutions found in the last step. (Enter your answers as a comma-separated list.)


Since we already found that the solution in the interval \([0, 2\pi)\) is \(\theta = \frac{\pi}{2}\), the general solution can be expressed by adding integer multiples of \(2\pi\) to this solution.

Let \(k\) be any integer. The general solutions for the equation \(\sin(\theta) = 1\) are:

\[
\theta = \frac{\pi}{2} + 2k\pi
\]

So, the complete set of solutions is:

\[
\theta = \frac{\pi}{2} + 2k\pi
\]

This represents all possible angles where \(\sin(\theta) = 1\).

Would you like more details or have any questions?

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Here are 8 related questions you might find useful:

1. How would you find the general solution for \(\cos(\theta) = 1\)?
2. What is the general solution for \(\sin(\theta) = -1\)?
3. How do you solve \(\sin(2\theta) = 1\) for all possible \(\theta\)?
4. Can you express the general solution for \(\sin(\theta) = 0.5\)?
5. How does the general solution change if the equation were \(\cos(\theta) = 1\)?
6. How do you find the general solutions for \(\tan(\theta) = 1\)?
7. What is the significance of the integer \(k\) in general solutions of trigonometric equations?
8. How would you solve \(\sin(\theta) = 0\) and express the general solution?

**Tip:** General solutions for trigonometric equations involve adding multiples of the function's period to the specific solutions, ensuring that all possible angles are covered.

Explore the general solution for the trigonometric equation sin(theta) = 1. Learn how to find all possible angles where sin(theta) equals 1 by adding integer multiples of 2pi to specific solutions. Suitable for advanced high school students and those studying trigonometry.

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