To solve the equation $\sin 2\theta = 1$ for $0 < \theta < 2\pi$, follow these steps:

Step 1: Find the general solution for $2\theta$

The sine function equals 1 at specific points: $\sin 2\theta = 1 \implies 2\theta = \frac{\pi}{2} + 2k\pi, \quad k \in \mathbb{Z}$ where $k$ is any integer.

Step 2: Solve for $\theta$

Divide both sides by 2 to solve for $\theta$: $\theta = \frac{\frac{\pi}{2} + 2k\pi}{2} = \frac{\pi}{4} + k\pi$

Step 3: Determine valid solutions within the given range

Now, we consider values of $\theta$ in the range $0 < \theta < 2\pi$. We test values of $k$:

1. For $k = 0$: $\theta = \frac{\pi}{4}$
2. For $k = 1$: $\theta = \frac{\pi}{4} + \pi = \frac{5\pi}{4}$
3. For $k = 2$: $\theta = \frac{\pi}{4} + 2\pi = \frac{9\pi}{4}$ However, $\frac{9\pi}{4}$ is greater than $2\pi$, so it is not within the range.

Step 4: List the solutions

The solutions for $\theta$ in the range $0 < \theta < 2\pi$ are: $\theta = \frac{\pi}{4}, \frac{5\pi}{4}$

These are the values of $\theta$ that satisfy the equation.

Would you like more details or have any questions?

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Here are 5 related questions:

1. What are the general solutions for $\sin x = -1$?
2. How does the range of the sine function affect the solutions of trigonometric equations?
3. Can you solve $\cos 2\theta = 1$ for the same interval?
4. How would the solution change if the equation was $\sin 2\theta = -1$?
5. What are the implications of solving $\sin 2\theta = \frac{1}{2}$ within the same interval?

Tip: Always check your solutions by substituting them back into the original equation to verify correctness.