Find real q such that `(3+2isintheta)/(1-2isintheta)` is purely real.

Explanation

To find the real \( q \) such that

\[ \frac{3 + 2i \sin \theta}{1 - 2i \sin \theta} \]

is purely real, we can follow these steps:

### Step 1: Rationalize the denominator
We will multiply the numerator and denominator by the conjugate of the denominator:

\[ \frac{3 + 2i \sin \theta}{1 - 2i \sin \theta} \cdot \frac{1 + 2i \sin \theta}{1 + 2i \sin \theta} \]

### Step 2: Simplify the expression
This gives us:

\[ \frac{(3 + 2i \sin \theta)(1 + 2i \sin \theta)}{(1 - 2i \sin \theta)(1 + 2i \sin \theta)} \]

The denominator simplifies to:

\[ 1^2 - (2i \sin \theta)^2 = 1 - 4(-\sin^2 \theta) = 1 + 4 \sin^2 \theta \]

The numerator simplifies to:

\[ 3(1) + 3(2i \sin \theta) + (2i \sin \theta)(1) + (2i \sin \theta)(2i \sin \theta) \]

Calculating this gives:

\[ 3 + 6i \sin \theta + 2i \sin \theta - 4 \sin^2 \theta = 3 - 4 \sin^2 \theta + 8i \sin \theta \]

### Step 3: Combine the results
Putting it all together, we have:

\[ \frac{3 - 4 \sin^2 \theta + 8i \sin \theta}{1 + 4 \sin^2 \theta} \]

### Step 4: Separate real and imaginary parts
Now we can separate the real and imaginary parts:

- Real part: \( \frac{3 - 4 \sin^2 \theta}{1 + 4 \sin^2 \theta} \)
- Imaginary part: \( \frac{8 \sin \theta}{1 + 4 \sin^2 \theta} \)

### Step 5: Set the imaginary part to zero
For the expression to be purely real, the imaginary part must be zero:

\[ \frac{8 \sin \theta}{1 + 4 \sin^2 \theta} = 0 \]

This implies:

\[ 8 \sin \theta = 0 \]

### Step 6: Solve for \( \sin \theta \)
Thus, we have:

\[ \sin \theta = 0 \]

The solutions for \( \theta \) are:

\[ \theta = n\pi \quad \text{where } n \in \mathbb{Z} \]

### Final Answer
The values of \( \theta \) that make the expression purely real are:

\[ \theta = n\pi \quad \text{for } n \in \mathbb{Z} \]

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