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

Answer

To find the real values of \( \theta \) such that \( \frac{3 + 2i \sin \theta}{1 - 2i \sin \theta} \) is purely real, we can follow these steps:

### Step 1: Write the Expression
We start with the expression:
\[ z = \frac{3 + 2i \sin \theta}{1 - 2i \sin \theta} \]

### Step 2: Multiply by the Conjugate
To eliminate the imaginary part in the denominator, we multiply the numerator and the denominator by the conjugate of the denominator:
\[ z = \frac{(3 + 2i \sin \theta)(1 + 2i \sin \theta)}{(1 - 2i \sin \theta)(1 + 2i \sin \theta)} \]

### Step 3: Simplify the Denominator
The denominator simplifies as follows:
\[ (1 - 2i \sin \theta)(1 + 2i \sin \theta) = 1^2 - (2i \sin \theta)^2 = 1 - 4(-1)(\sin^2 \theta) = 1 + 4 \sin^2 \theta \]

### Step 4: Simplify the Numerator
Now, we simplify the numerator:
\[ (3 + 2i \sin \theta)(1 + 2i \sin \theta) = 3 + 6i \sin \theta + 2i \sin \theta + 4(-1)(\sin^2 \theta) = 3 + (6i + 2i) \sin \theta - 4 \sin^2 \theta = 3 - 4 \sin^2 \theta + 8i \sin \theta \]

### Step 5: Combine the Results
Now we can combine both parts:
\[ z = \frac{3 - 4 \sin^2 \theta + 8i \sin \theta}{1 + 4 \sin^2 \theta} \]

### Step 6: Separate Real and Imaginary Parts
We can separate the real and imaginary parts:
\[ z = \frac{3 - 4 \sin^2 \theta}{1 + 4 \sin^2 \theta} + i \frac{8 \sin \theta}{1 + 4 \sin^2 \theta} \]

### Step 7: Set the Imaginary Part to Zero
For \( z \) to be purely real, the imaginary part must be zero:
\[ \frac{8 \sin \theta}{1 + 4 \sin^2 \theta} = 0 \]

### Step 8: Solve for \( \sin \theta \)
This implies:
\[ 8 \sin \theta = 0 \quad \Rightarrow \quad \sin \theta = 0 \]

### Step 9: Find Values of \( \theta \)
The solutions for \( \sin \theta = 0 \) are:
\[ \theta = n\pi \quad \text{where } n \in \mathbb{Z} \]

### Conclusion
Thus, the required values of \( \theta \) such that the expression is purely real are:
\[ \theta = n\pi \quad \text{for any integer } n \]