Let A = left{ frac{1967 + 1686 i sin theta}{7 | vedclass

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Question

(A) Given $A = \frac{1967 + 1686 i \sin 	heta}{7 - 3 i \cos 	heta}$.We can factor out $281$ from the numerator: $A = \frac{281(7 + 6 i \sin 	heta)}

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$281$

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(A) Given $A = \frac{1967 + 1686 i \sin 	heta}{7 - 3 i \cos 	heta}$.We can factor out $281$ from the numerator: $A = \frac{281(7 + 6 i \sin 	heta)}{7 - 3 i \cos 	heta}$.To simplify,multiply the numerator and denominator by the conjugate of the denominator $(7 + 3 i \cos 	heta)$:$A = \frac{281(7 + 6 i \sin 	heta)(7 + 3 i \cos 	heta)}{(7 - 3 i \cos 	heta)(7 + 3 i \cos 	heta)} = \frac{281(49 + 21 i \cos 	heta + 42 i \sin 	heta - 18 \sin 	heta \cos 	heta)}{49 + 9 \cos^2 	heta}$.For $A$ to be a real number,the imaginary part must be zero:$21 \cos 	heta + 42 \sin 	heta = 0 \implies 	an 	heta = -\frac{1}{2}$.Using $	an 	heta = -\frac{1}{2}$,we find $\sin 2 	heta = \frac{2 	an 	heta}{1 + 	an^2 	heta} = \frac{-1}{1 + 1/4} = -\frac{4}{5}$ and $\cos^2 	heta = \frac{1}{1 + 	an^2 	heta} = \frac{1}{1 + 1/4} = \frac{4}{5}$.Substituting these into the real part of $A$:$A = \frac{281(49 - 9 \sin 2 	heta)}{49 + 9 \cos^2 	heta} = \frac{281(49 - 9(-4/5))}{49 + 9(4/5)} = \frac{281(49 + 36/5)}{49 + 36/5} = 281$.Thus,the only positive integer $n$ is $281$.

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

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