Find the `theta` such that `(3+2i sin theta)/(1-2 isin theta)` is
(a) real

Answer

To find the values of \( \theta \) such that the expression

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

is real, we need to ensure that the imaginary part of this complex number is equal to zero.

### Step-by-step Solution:

**Step 1: Identify the imaginary part.**

The expression is in the form of a complex number \( \frac{a + bi}{c + di} \). We need to find the imaginary part of this fraction.

**Step 2: Multiply by the conjugate.**

To simplify, we multiply the numerator and the denominator by the conjugate of the denominator:

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

**Step 3: Simplify the denominator.**

The denominator becomes:

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

**Step 4: Simplify the numerator.**

Now, simplify the numerator:

\[ (3 + 2i \sin \theta)(1 + 2i \sin \theta) = 3 + 6i \sin \theta + 2i \sin \theta + 4(-\sin^2 \theta) = 3 - 4 \sin^2 \theta + 8i \sin \theta \]

**Step 5: Combine the results.**

Putting it all together, we have:

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

**Step 6: Identify the imaginary part.**

The imaginary part of this expression is:

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

**Step 7: Set the imaginary part to zero.**

For the expression to be real, we set the imaginary part to zero:

\[ 8 \sin \theta = 0 \]

**Step 8: Solve for \( \theta \).**

This implies:

\[ \sin \theta = 0 \]

The general solution for \( \sin \theta = 0 \) is:

\[ \theta = n\pi \quad \text{where } n \text{ is an integer.} \]

### Final Answer:
The values of \( \theta \) such that \( \frac{3 + 2i \sin \theta}{1 - 2i \sin \theta} \) is real are:

\[ \theta = n\pi \quad (n \in \mathbb{Z}) \]