Let $A=\left\{\dfrac{1967+1686i\sin\theta}{7-3i\cos\theta}:\theta\in\mathbb{R}\right\}$. If $A$ contains exactly one positive integer $n$, then the value of $n$ is ___.
Step-by-Step Solution
Key Concept: Set imaginary part of z to zero to find θ; then compute the resulting real value
Let $z=\dfrac{1967+1686i\sin\theta}{7-3i\cos\theta}$. Multiply numerator and denominator by $\overline{(7-3i\cos\theta)}=7+3i\cos\theta$:
Imaginary part of numerator: $1967\cdot3\cos\theta+1686\sin\theta\cdot7=5901\cos\theta+11802\sin\theta=5901(\cos\theta+2\sin\theta)$.
For $z$ real: $\cos\theta+2\sin\theta=0\Rightarrow\tan\theta=-1/2$. Taking $\sin\theta=-1/\sqrt5,\cos\theta=2/\sqrt5$:
Real part $= \dfrac{1967\cdot7+1686\cdot(-1/\sqrt5)\cdot(-3)\cdot... }{...}$
With $\sin\theta\cos\theta=-2/5$: numerator real part $=1967\cdot7-1686\cdot3\cdot(-2/5)\cdot...$ Let me use:
Real numerator $=13769-5058\sin\theta\cos\theta=13769-5058(-2/5)=13769+\dfrac{10116}{5}=\dfrac{78961}{5}$.
Denominator $=49+9\cos^2\theta=49+\dfrac{36}{5}=\dfrac{281}{5}$.
$z=\dfrac{78961/5}{281/5}=\dfrac{78961}{281}=281$.
For both choices of $\tan\theta=-1/2$, the result is $281$. So the only real value in $A$ is $281$, the only positive integer. $n=281$.
Correct Answer: 281