Let $A=\left\{\frac{1967+1686 i \sin heta}{7-3 i \cos heta}: heta \in \mathbb{R} ight\}$. If

Question

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

Professor PC Thomas Classes · Answer

(281) : $A=\frac{1967+1686 i \sin 	heta}{7-3 i \cos 	heta}$\ $=\frac{281(7+6 i \sin 	heta)}{7-3 i \cos 	heta} 	imes \frac{7+3 i \cos 	heta}{7+3 i \cos 	heta}$\ $=\frac{281(49-9 \sin 2 	heta)}{49+9 \cos ^{2} 	heta}+\frac{5901 i(2 \sin 	heta+\cos 	heta)}{49+9 \cos ^{2} 	heta}$\ For it to be positive integer (i.e., real number) $$ 2 \sin 	heta+\cos 	heta=0 $$ $\Rightarrow 	an 	heta=-\frac{1}{2}, \sin 	heta=\frac{1}{\sqrt{5}}$ and $\cos 	heta=\frac{-2}{\sqrt{5}}$\ $	herefore \quad \frac{281\left(49+18\left(\frac{1}{\sqrt{5}}ight)\left(\frac{2}{\sqrt{5}}ight)ight)}{49+9(415)}=218$\