Value of Positive Integer in Complex Number Set

To find the positive integer $n \in A$, we set the given complex expression equal to $n$:

$n = \frac{1967 + 1686i \sin \theta}{7 - 3i \cos \theta}$

Multiplying both the numerator and the denominator by the complex conjugate of the denominator, $7 + 3i \cos \theta$, we get:

$n = \frac{(1967 + 1686i \sin \theta)(7 + 3i \cos \theta)}{(7 - 3i \cos \theta)(7 + 3i \cos \theta)}$

Expanding the terms in the numerator:

$n = \frac{(1967 \cdot 7 - 1686 \cdot 3 \sin \theta \cos \theta) + i(1967 \cdot 3 \cos \theta + 1686 \cdot 7 \sin \theta)}{49 + 9 \cos^2 \theta}$

Since $n$ is a real number (a positive integer), the imaginary part of this expression must be equal to zero:

$\text{Im}(n) = \frac{1967 \cdot 3 \cos \theta + 1686 \cdot 7 \sin \theta}{49 + 9 \cos^2 \theta} = 0$

Equating the numerator of the imaginary part to zero:

$5901 \cos \theta + 11802 \sin \theta = 0$

Dividing the entire equation by $5901$:

$\cos \theta + 2 \sin \theta = 0 \implies \tan \theta = -\frac{1}{2}$

From $\tan \theta = -\frac{1}{2}$, we can calculate:

$\sin^2 \theta = \frac{\tan^2 \theta}{1 + \tan^2 \theta} = \frac{1/4}{5/4} = \frac{1}{5}$

$\cos^2 \theta = \frac{1}{1 + \tan^2 \theta} = \frac{1}{5/4} = \frac{4}{5}$

$\sin \theta \cos \theta = \sin \theta (-2 \sin \theta) = -2 \sin^2 \theta = -\frac{2}{5}$

Now, substituting these values into the real part to find $n$:

$n = \frac{1967 \cdot 7 - 1686 \cdot 3 \sin \theta \cos \theta}{49 + 9 \cos^2 \theta}$

$n = \frac{13769 - 5058 \left(-\frac{2}{5}\right)}{49 + 9 \left(\frac{4}{5}\right)}$

$n = \frac{13769 + \frac{10116}{5}}{49 + \frac{36}{5}}$

Multiplying both numerator and denominator by $5$:

$n = \frac{5 \cdot 13769 + 10116}{5 \cdot 49 + 36} = \frac{68845 + 10116}{245 + 36}$

$n = \frac{78961}{281}$

Since $281 \times 281 = 78961$:

$n = 281$

Thus, the value of the positive integer $n$ is $281$.