Let A $= \{\frac{1967 + 1686 i \sinθ}{7 - 3 i \cosθ} : θ ∈ R\}.$ If A contains exactly one positive integer n, then the value of n is __________.

Let A={(1967+1686 isinθ)/(7-3 icosθ):θ∈R}. If A contains exactly one positive integer n, then the value of n is

$A=\Big\{\dfrac{1967+1686\,i\sin\theta}{7-3\,i\cos\theta}:\theta\in\mathbb{R}\Big\}$

$\text{Given: } A \text{ has exactly one positive integer } n.$

$\text{To find: } n$

$1.\ \text{split } z \text{ into real and imaginary parts}$

$2.\ \text{set } \operatorname{Im} z = 0 \text{ and solve for } \theta$

$3.\ \text{the real part at that } \theta \text{ is the positive integer } n$

$1967 = 7\times 281,\qquad 1686 = 6\times 281$

$z = 281\cdot\dfrac{7+6\,i\sin\theta}{7-3\,i\cos\theta}$

$\dfrac{7+6\,i\sin\theta}{7-3\,i\cos\theta}=\dfrac{(7+6\,i\sin\theta)(7+3\,i\cos\theta)}{49+9\cos^2\theta}$

$=\dfrac{49-18\sin\theta\cos\theta+i\,(21\cos\theta+42\sin\theta)}{49+9\cos^2\theta}$

$\operatorname{Re}z=281\cdot\dfrac{49-18\sin\theta\cos\theta}{49+9\cos^2\theta}$

$\operatorname{Im}z=281\cdot\dfrac{21\cos\theta+42\sin\theta}{49+9\cos^2\theta}$

$\operatorname{Im}z=0\ \Rightarrow\ 21\cos\theta+42\sin\theta=0$

$\Rightarrow\ \cos\theta+2\sin\theta=0\ \Rightarrow\ \tan\theta=-\dfrac{1}{2}$

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

$\sin\theta\cos\theta=\tan\theta\cdot\cos^2\theta=-\dfrac{1}{2}\cdot\dfrac{4}{5}=-\dfrac{2}{5}$

$\operatorname{Re}z=281\cdot\dfrac{49-18\left(-\frac{2}{5}\right)}{49+9\left(\frac{4}{5}\right)}$

$=281\cdot\dfrac{49+\frac{36}{5}}{49+\frac{36}{5}}=281$

$n=281$