Question

Let A = {$\frac{1967+1686 i\,sin\theta}{7-3i\,cos\theta}:\theta\in R$ }. If A contains exactly one positive integer n, then the value of n is ______

Answer 1

So, let's write the numbers i.e (1967, 1686) as factors of 7 and 3
1967 = 7 × 281 and 1686 = 6 × 562
So, the expression will be as follows :
$\frac{281(7)+281(6i\sin(x))}{7-3i(\cos(x))}$ $⇒281(\frac{7+6i\sin(x)}{7-3i\cos(x)})$ …. (i)
Now, by multiplying both numerator and denominator by $7 + 3i\sin(x)$, we get :
$=281(\frac{49+21i(\cos(x)+2\sin(x))+18\sin(x)\cos(x)}{49+9\cos^2(x)})$
$⇒281(\frac{49+18\sin(x)\cos(x)}{49+9\cos^2}+i\frac{21(\cos(x)+2(x)\sin)}{49+9\cos^2})$
Since we know the expression will be an integer after simplification, we can conclude that the imaginary part of the expression is zero.
Therefore, we can say that
$(\frac{21(\cos(x)+2\sin(x))}{49+9\cos^2})=0$
After further simplification, it turns out to be $\cos(x)=-2\sin(x)$.
Now, substitute the value of cos(x) back into the equation, and the equation becomes
$⇒281(\frac{7+6i\sin(x)}{7+6i\sin(x)})=281$
Therefore, the value of n is 281.