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Step-by-Step Solution
Step 1: Write the Given Complex Expression
We have the complex expression
$$
\frac{3 + 2\,i\,\sin \theta}{1 - 2\,i\,\sin \theta},
$$
and we need to determine for which values of $ \theta $ (in the interval $ -\frac{\pi}{2} < \theta < \pi$) this expression is purely imaginary.
Step 2: Multiply Numerator and Denominator by the Conjugate of the Denominator
The denominator is $1 - 2\,i\,\sin\theta$. Its conjugate is $1 + 2\,i\,\sin\theta$. So we multiply the numerator and denominator by $1 + 2\,i\,\sin\theta$:
$$
\frac{3 + 2\,i\,\sin \theta}{1 - 2\,i\,\sin \theta}
\times
\frac{1 + 2\,i\,\sin \theta}{1 + 2\,i\,\sin \theta}
=
\frac{(3 + 2\,i\,\sin \theta)\,(1 + 2\,i\,\sin \theta)}{(1 - 2\,i\,\sin \theta)\,(1 + 2\,i\,\sin \theta)}.
$$
Step 3: Expand the Numerator and the Denominator
• Numerator:
$$
(3 + 2\,i\,\sin \theta)\,(1 + 2\,i\,\sin \theta)
= 3(1) + 3(2\,i\,\sin \theta) + 2\,i\,\sin \theta \cdot 1 + 2\,i\,\sin \theta \cdot 2\,i\,\sin \theta.
$$
This simplifies to
$$
3 + 6\,i\,\sin \theta + 2\,i\,\sin \theta + 4\,i^2\,(\sin \theta)^2.
$$
Since $i^2 = -1$,
$$
4\,i^2\,(\sin \theta)^2 = 4(-1)(\sin \theta)^2 = -4\,(\sin \theta)^2.
$$
Thus, the numerator becomes
$$
(3 - 4\,(\sin \theta)^2) + i\,\bigl(6\,\sin \theta + 2\,\sin \theta\bigr)
= \bigl(3 - 4\,(\sin \theta)^2\bigr) + i\,\bigl(8\,\sin \theta\bigr).
$$
• Denominator:
$$
(1 - 2\,i\,\sin \theta)\,(1 + 2\,i\,\sin \theta)
= 1^2 - (2\,i\,\sin\theta)^2
= 1 - 4\,i^2\,(\sin \theta)^2
= 1 + 4\,(\sin \theta)^2
$$
(using $i^2 = -1$ again).
Hence, the expression becomes
$$
\frac{\bigl(3 - 4\,(\sin \theta)^2\bigr) + i\,\bigl(8\,\sin \theta\bigr)}{1 + 4\,(\sin \theta)^2}.
$$
Step 4: Impose the Purely Imaginary Condition
For the entire fraction to be purely imaginary, its real part must be zero. The real part of
$$
\frac{\bigl(3 - 4\,(\sin \theta)^2\bigr) + i\,\bigl(8\,\sin \theta\bigr)}{1 + 4\,(\sin \theta)^2}
$$
is
$$
\frac{3 - 4\,(\sin \theta)^2}{1 + 4\,(\sin \theta)^2}.
$$
Setting this equal to zero:
$$
\frac{3 - 4\,(\sin \theta)^2}{1 + 4\,(\sin \theta)^2} = 0
\quad \Longrightarrow \quad
3 - 4\,(\sin \theta)^2 = 0.
$$
Step 5: Solve for $ \theta $
From $3 - 4\,(\sin \theta)^2 = 0$, we get
$$
4\,(\sin \theta)^2 = 3
\quad \Longrightarrow \quad
(\sin \theta)^2 = \frac{3}{4}
\quad \Longrightarrow \quad
\sin \theta = \pm \frac{\sqrt{3}}{2}.
$$
Step 6: Identify Valid Angles in the Given Interval
We have $ -\frac{\pi}{2} < \theta < \pi$. The angles that satisfy
$ \sin \theta = \frac{\sqrt{3}}{2} $ in this interval are
$ \theta = \frac{\pi}{3}, \frac{2\pi}{3}. $
The angle that satisfies
$ \sin \theta = -\frac{\sqrt{3}}{2} $
in this interval is
$ \theta = -\frac{\pi}{3} $
(since $-\frac{\pi}{2}< -\frac{\pi}{3} < 0$).
Therefore, the three valid solutions are
$$
\theta = -\frac{\pi}{3}, \quad \theta = \frac{\pi}{3}, \quad \theta = \frac{2\pi}{3}.
$$
Step 7: Sum the Solutions
We now sum these solutions:
$$
\left(-\frac{\pi}{3}\right) + \left(\frac{\pi}{3}\right) + \left(\frac{2\pi}{3}\right)
= 0 + \frac{2\pi}{3}
= \frac{2\pi}{3}.
$$
Final Answer
The sum of the elements of set $A$ is
$$
\frac{2\pi}{3}.
$$
