Let $z$ be an element of the set $A$. The expression for $z$ is given by: $z = \frac{1967+1686 i \sin \theta}{7-3 i \cos \theta}, \quad \theta \in \mathbb{R}$

The problem states that the set $A$ contains exactly one positive integer, which we'll call $n$. This means that for some value of $\theta$, say $\theta_0$, the value of $z$ is equal to $n$. $n = \frac{1967+1686 i \sin \theta_0}{7-3 i \cos \theta_0}$

Step 1: Condition for $z$ to be a real number

For $z$ to be a real number $n$, its imaginary part must be zero. To find the imaginary part, we rationalize the denominator of the expression for $z$: $z = \frac{1967+1686 i \sin \theta}{7-3 i \cos \theta} \times \frac{7+3 i \cos \theta}{7+3 i \cos \theta}$ $z = \frac{(1967)(7) + (1967)(3i \cos \theta) + (1686 i \sin \theta)(7) + (1686 i \sin \theta)(3i \cos \theta)}{7^2 - (3i \cos \theta)^2}$ $z = \frac{13769 + 5901 i \cos \theta + 11802 i \sin \theta - 5058 \sin \theta \cos \theta}{49 + 9 \cos^2 \theta}$ $z = \frac{(13769 - 5058 \sin \theta \cos \theta) + i(5901 \cos \theta + 11802 \sin \theta)}{49 + 9 \cos^2 \theta}$ For $z$ to be real, the imaginary part must be zero: $\text{Im}(z) = \frac{5901 \cos \theta + 11802 \sin \theta}{49 + 9 \cos^2 \theta} = 0$ Since the denominator $49 + 9 \cos^2 \theta$ is always positive, the numerator must be zero: $5901 \cos \theta + 11802 \sin \theta = 0$ Noticing that $11802 = 2 \times 5901$, we can simplify the equation: $5901 (\cos \theta + 2 \sin \theta) = 0$ $\cos \theta + 2 \sin \theta = 0$ $2 \sin \theta = -\cos \theta$ $\tan \theta = -\frac{1}{2}$ This condition tells us that for any $\theta$ satisfying $\tan\theta = -1/2$, the corresponding element $z$ of set $A$ is a real number.

Step 2: Finding the value of the integer $n$

Now we find the value of this real number, which must be our integer $n$. The value of $n$ is the real part of $z$ when the imaginary part is zero: $n = \text{Re}(z) = \frac{13769 - 5058 \sin \theta \cos \theta}{49 + 9 \cos^2 \theta}$ From the condition $\tan \theta = -1/2$, we can determine the values of $\sin\theta\cos\theta$ and $\cos^2\theta$. We can visualize a right-angled triangle with opposite side 1 and adjacent side 2. The hypotenuse would be $\sqrt{1^2+2^2} = \sqrt{5}$. Since $\tan\theta$ is negative, $\theta$ is in the second or fourth quadrant.

• If $\theta$ is in the second quadrant: $\sin \theta = \frac{1}{\sqrt{5}}$ and $\cos \theta = -\frac{2}{\sqrt{5}}$.
• If $\theta$ is in the fourth quadrant: $\sin \theta = -\frac{1}{\sqrt{5}}$ and $\cos \theta = \frac{2}{\sqrt{5}}$. In both cases, we have: $\sin \theta \cos \theta = \left(\pm\frac{1}{\sqrt{5}}\right)\left(\mp\frac{2}{\sqrt{5}}\right) = -\frac{2}{5}$ $\cos^2 \theta = \left(\pm\frac{2}{\sqrt{5}}\right)^2 = \frac{4}{5}$ Substitute these values into the expression for $n$: $n = \frac{13769 - 5058 \left(-\frac{2}{5}\right)}{49 + 9 \left(\frac{4}{5}\right)} = \frac{13769 + \frac{10116}{5}}{49 + \frac{36}{5}}$ To simplify the fraction, multiply the numerator and denominator by 5: $n = \frac{13769 \times 5 + 10116}{49 \times 5 + 36} = \frac{68845 + 10116}{245 + 36} = \frac{78961}{281}$ Performing the division: $n = 281$

Step 3: Conclusion

We have found that the only real value that any element in set $A$ can take is 281. Since 281 is a positive integer, it is the unique positive integer $n$ in the set $A$. The condition that $A$ contains exactly one positive integer is satisfied.

Alternative Method

Notice that the coefficients in the numerator and denominator are related: $1967 = 7 \times 281$ and $1686 = 6 \times 281$. $z = \frac{281(7 + 6i \sin \theta)}{7 - 3i \cos \theta} = 281 \left( \frac{7 + 6i \sin \theta}{7 - 3i \cos \theta} \right)$ For $z$ to be a real number $n$, the complex fraction must evaluate to a real number $k$, such that $n = 281k$. $k = \frac{7 + 6i \sin \theta}{7 - 3i \cos \theta}$. For $k$ to be real, the imaginary part of its rationalized form must be zero. $k = \frac{(7+6 i \sin \theta)(7+3 i \cos \theta)}{49+9\cos^2\theta} = \frac{(49-18\sin\theta\cos\theta)+i(21\cos\theta+42\sin\theta)}{49+9\cos^2\theta}$ Setting the imaginary part to zero gives $21\cos\theta+42\sin\theta=0$, which simplifies to $\tan\theta = -1/2$, as before. For this value of $\theta$, $k$ is the real part: $k = \frac{49-18\sin\theta\cos\theta}{49+9\cos^2\theta} = \frac{49-18(-2/5)}{49+9(4/5)} = \frac{49+36/5}{49+36/5} = 1$ Therefore, $n = 281 \times k = 281 \times 1 = 281$.

The value of $n$ is 281.