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We are given the equation:

\[
\frac{3 + 2i}{1 - 2i \sin \theta}
\]
is a real number, and \( 0 \leq \theta \leq 2\pi \).

We need to determine the value of \( \theta \) such that the expression is real.

### Step-by-step solution:

For the given complex number to be real, the imaginary part of the quotient must be 0.

Let’s consider the expression:

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

We will multiply both the numerator and the denominator by the conjugate of the denominator to simplify the expression. The conjugate of \( 1 - 2i \sin \theta \) is \( 1 + 2i \sin \theta \). Thus, we multiply:

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

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

### Numerator:
Expanding \( (3 + 2i)(1 + 2i \sin \theta) \):
\[
= 3(1 + 2i \sin \theta) + 2i(1 + 2i \sin \theta)
\]
\[
= 3 + 6i \sin \theta + 2i + 4\sin \theta
\]
\[
= (3 + 4\sin \theta) + i(6\sin \theta + 2)
\]

Thus, the full expression becomes:
\[
\frac{(3 + 4\sin \theta) + i(6\sin \theta + 2)}{1 - 4\sin^2 \theta}
\]

### Condition for Real Number:
For the expression to be a real number, the imaginary part must be zero. Therefore, we need:

\[
6\sin \theta + 2 = 0
\]

Solving for \( \sin \theta \):
\[
6\sin \theta = -2
\]
\[
\sin \theta = -\frac{1}{3}
\]

Now, we need to find the values of \( \theta \) for which \( \sin \theta = -\frac{1}{3} \), given that \( 0 \leq \theta \leq 2\pi \).

### Solution for \( \theta \):
The solutions for \( \theta \) where \( \sin \theta = -\frac{1}{3} \) occur in the third and fourth quadrants. Using the inverse sine function:

\[
\theta = \pi + \arcsin\left(-\frac{1}{3}\right) \quad \text{(third quadrant)}
\]
\[
\theta = 2\pi - \arcsin\left(-\frac{1}{3}\right) \quad \text{(fourth quadrant)}
\]

Approximating \( \arcsin\left(\frac{1}{3}\right) \), we get \( \theta \approx \frac{7\pi}{6} \) or \( \theta \approx \frac{5\pi}{3} \).

Among the provided options, the closest solution is:
\[
\boxed{\frac{5\pi}{3}}
\]

Would you like more details or clarifications?

### Related Questions:
1. How do you find the conjugate of a complex number?
2. What is the importance of making the imaginary part zero in such problems?
3. How does multiplying by the conjugate simplify complex fractions?
4. How do we solve trigonometric equations involving sine?
5. What are the general solutions for sine in different quadrants?

### Tip:
When simplifying complex fractions, multiplying by the conjugate is a powerful method to remove imaginary terms in the denominator.

If (3 + 2i) / (1 - 2i sin(θ)) is a real number and 0 ≤ θ ≤ 2π, then θ = ...

Learn how to solve for θ in the complex equation (3 + 2i) / (1 - 2i sin(θ)) where the result must be a real number. This problem involves multiplying complex numbers by their conjugates and solving trigonometric equations. Suitable for students in Grades 10-12.

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