Let $A=\left\{\frac{1967+1686isin\theta }{7-3icos\theta }:\theta \in ℝ\right)$.

If $A$ contains exactly one positive integer $n$, then the value of $n$ is

`. Let's check the MathML tags carefully: `

The correct answer is 281.

` Let's construct the MathML representation step-by-step: Paragraph 1: `

The correct answer is 281.

` Paragraph 2: `

We are given the set:
$A=\left\{\frac{1967+1686isin\theta }{7-3icos\theta }:\theta \in ℝ\right\}$
We are told that $A$ contains exactly one positive integer $n$. Since $n$ is a real integer, the complex number in $A$ must be purely real for some value of $\theta \in ℝ$.

` Paragraph 3: `

Let $z=\frac{1967+1686isin\theta }{7-3icos\theta }=n$.
Multiplying both sides by the denominator, we get:
$1967+1686isin\theta =n\left(7-3icos\theta \right)$
$1967+1686isin\theta =7n-3nicos\theta$

` Paragraph 4: `

Equating the real and imaginary parts from both sides:
1. Real parts:
$1967=7n$
$n=\frac{1967}{7}=281$

` Paragraph 5: `

2. Imaginary parts:
$1686sin\theta =-3ncos\theta$
Substituting $n=281$ into the imaginary equation:
$1686sin\theta =-3\left(281\right)cos\theta$
$1686sin\theta =-843cos\theta$
$tan\theta =-\frac{843}{1686}=-\frac{1}{2}$

` Paragraph 6: `

Since $tan\theta =-\frac{1}{2}$ has real solutions for $\theta \in ℝ$, there exists a real angle $\theta$ such that the complex number simplifies to the positive integer $n=281$.

` Paragraph 7: `

Thus, the value of $n$ is 281.

` Let's double-check all requirements: - Clean core answer: `281` - MathML used correctly, no LaTeX (`$..$`). - Raw unicode `-`, `+`, `=`, `⇒` (no HTML entities for symbols like `−`). - Every `$`\; tag\; line\; has\; `$
` before/after or is in its own `

`. - No `display="block"` anywhere. Everything looks completely solid! 281

The correct answer is 281.

We are given the set:
$A=\left\{\frac{1967+1686isin\theta }{7-3icos\theta }:\theta \in ℝ\right\}$
We are told that $A$ contains exactly one positive integer $n$. For a element in $A$ to be a positive integer (and thus a real number), its value must equal $n$ for some real angle $\theta$.

Let:
$\frac{1967+1686isin\theta }{7-3icos\theta }=n$
Multiplying both sides by the denominator:
$1967+1686isin\theta =n\left(7-3icos\theta \right)$
Expanding the right-hand side:
$1967+1686isin\theta =7n-3nicos\theta$

Equating the real and imaginary components from both sides of the equation:

1. Real Parts:
$1967=7n$
Solving for $n$:
$n=\frac{1967}{7}=281$

2. Imaginary Parts:
$1686sin\theta =-3ncos\theta$
Substitute $n=281$ into this relation:
$1686sin\theta =-3\left(281\right)cos\theta$
$1686sin\theta =-843cos\theta$
Rearranging to solve for $tan\theta$:
$tan\theta =-\frac{843}{1686}=-\frac{1}{2}$

Since $tan\theta =-\frac{1}{2}$ yields valid real solutions for $\theta \in ℝ$, there exists a real value of $\theta$ for which the given complex expression evaluates precisely to the positive integer $n=281$.

Therefore, the value of $n$ is 281.