list six solutions for sin theta over 2=root 2 over 2 

We are given the equation:

\[
\sin\left(\frac{\theta}{2}\right) = \frac{\sqrt{2}}{2}
\]

The value \(\frac{\sqrt{2}}{2}\) corresponds to an angle of \(45^\circ\) or \(\frac{\pi}{4}\) radians in the unit circle. Therefore, we will solve for \(\frac{\theta}{2}\) and then solve for \(\theta\).

### Step 1: Find solutions for \(\frac{\theta}{2}\)

Since \(\sin\) is positive in the first and second quadrants, the solutions for \(\frac{\theta}{2}\) are:

\[
\frac{\theta}{2} = \frac{\pi}{4} \quad \text{or} \quad \frac{\theta}{2} = \pi - \frac{\pi}{4} = \frac{3\pi}{4}
\]

### Step 2: General solutions for \(\frac{\theta}{2}\)

The general solutions for \(\frac{\theta}{2}\) are:

\[
\frac{\theta}{2} = \frac{\pi}{4} + 2k\pi \quad \text{or} \quad \frac{\theta}{2} = \frac{3\pi}{4} + 2k\pi \quad \text{for integers} \ k
\]

### Step 3: Solve for \(\theta\)

Now, multiplying both sides by 2 to solve for \(\theta\), we get:

\[
\theta = 2\left(\frac{\pi}{4} + 2k\pi\right) = \frac{\pi}{2} + 4k\pi
\]
\[
\theta = 2\left(\frac{3\pi}{4} + 2k\pi\right) = \frac{3\pi}{2} + 4k\pi
\]

### Step 4: List six solutions

To find six specific solutions, we can set \(k = 0, k = 1, k = -1\), etc.:

1. For \(k = 0\): \(\theta = \frac{\pi}{2}\) and \(\theta = \frac{3\pi}{2}\)
2. For \(k = 1\): \(\theta = \frac{\pi}{2} + 4\pi = \frac{9\pi}{2}\) and \(\theta = \frac{3\pi}{2} + 4\pi = \frac{11\pi}{2}\)
3. For \(k = -1\): \(\theta = \frac{\pi}{2} - 4\pi = \frac{-7\pi}{2}\) and \(\theta = \frac{3\pi}{2} - 4\pi = \frac{-5\pi}{2}\)

Thus, six solutions for \(\theta\) are:

\[
\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{9\pi}{2}, \frac{11\pi}{2}, \frac{-7\pi}{2}, \frac{-5\pi}{2}
\]

Would you like more details or clarifications on any of these steps?

### Related Questions:
1. What are the general solutions for \(\sin\theta = -\frac{\sqrt{2}}{2}\)?
2. How would the solutions change if \(\sin\left(\frac{\theta}{2}\right)\) were negative?
3. Can you explain how the sine function behaves on the unit circle?
4. What are the general solutions for \(\cos\theta = \frac{\sqrt{2}}{2}\)?
5. How would you find solutions for \(\tan\left(\frac{\theta}{2}\right) = \frac{1}{\sqrt{2}}\)?

### Tip:
Always consider the periodicity of trigonometric functions when solving equations.

list six solutions for sin(theta/2) = sqrt(2)/2

Find six specific solutions for the trigonometric equation sin(θ/2) = √2/2. This solution explores the sine function, unit circle properties, and periodicity to provide exact answers for θ. Suitable for high school students studying trigonometry.

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