If frac{3+2i sin theta}{1-2i sin theta} is a | vedclass

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(A) Let $z = \frac{3+2i \sin 	heta}{1-2i \sin 	heta}$. To simplify,multiply the numerator and denominator by the conjugate of the denominator $(1+2i

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$\pi$

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(A) Let $z = \frac{3+2i \sin 	heta}{1-2i \sin 	heta}$. To simplify,multiply the numerator and denominator by the conjugate of the denominator $(1+2i \sin 	heta)$:$z = \frac{(3+2i \sin 	heta)(1+2i \sin 	heta)}{(1-2i \sin 	heta)(1+2i \sin 	heta)}$$z = \frac{3 + 6i \sin 	heta + 2i \sin 	heta + 4i^2 \sin^2 	heta}{1^2 + (2 \sin 	heta)^2}$Since $i^2 = -1$,we have:$z = \frac{3 - 4 \sin^2 	heta + 8i \sin 	heta}{1 + 4 \sin^2 	heta}$$z = \frac{3 - 4 \sin^2 	heta}{1 + 4 \sin^2 	heta} + i \left( \frac{8 \sin 	heta}{1 + 4 \sin^2 	heta} ight)$For $z$ to be a real number,the imaginary part must be zero:$\frac{8 \sin 	heta}{1 + 4 \sin^2 	heta} = 0$This implies $\sin 	heta = 0$.Given $0 < 	heta < 2\pi$,the only solution is $	heta = \pi$.

If $\frac{3+2i \sin \theta}{1-2i \sin \theta}$ is a real number and $0 < \theta < 2\pi$,then $\theta$ is equal to

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