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Question: If \(\frac{3 + 2i\sin\theta}{1 - 2i\sin\theta}\) and \(\theta =\), then \(2n\pi \pm \frac{\pi}{3}\)....

If 3+2isin⁡θ1−2isin⁡θ\frac{3 + 2i\sin\theta}{1 - 2i\sin\theta} and θ=\theta =, then 2nπ±π32n\pi \pm \frac{\pi}{3}.

A

1

B

2

C

3

D

4

Answer

4

Explanation

Solution

If x2+y2=9(5+4cos⁡θ)2x^{2} + y^{2} = \frac{9}{(5 + 4\cos\theta)^{2}}and [4+cos⁡2θ+4cos⁡θ+sin⁡2θ]\lbrack 4 + \cos^{2}\theta + 4\cos\theta + \sin^{2}\theta\rbrack

Then =95+4cos⁡θ=4[6+3cos⁡θ5+4cos⁡θ]−3=4x−3= \frac{9}{5 + 4\cos\theta} = 4\left\lbrack \frac{6 + 3\cos\theta}{5 + 4\cos\theta} \right\rbrack - 3 = 4x - 3⇒ x+iy=3(2+cos⁡θ−isin⁡θ)(2+cos⁡θ+isin⁡θ)(2+cos⁡θ−isin⁡θ)x + iy = \frac{3(2 + \cos\theta - i\sin\theta)}{(2 + \cos\theta + i\sin\theta)(2 + \cos\theta - i\sin\theta)}.