A) \[\frac{5\pi }{6}\]
B) \[\frac{3\pi }{4}\]
C) \[\frac{2\pi }{3}\]
D) \[\pi \]
Correct Answer: C
Solution :
Given \[\theta \,\in \,\left( -\frac{\pi }{2},\,\,\pi \right)\] Let \[f(\theta )\,\,=\,\,\frac{3+2i\,\,\sin \,\theta }{1-2i\,\,\sin \,\theta }\] \[\,\,=\,\,\frac{(3+2i\,\,\sin \,\theta )(1+2i\,\,sin\,\theta )}{(1-2i\,\,\sin \,\theta )(1+2i\,\,\sin \,\,\theta )}\] \[f(\theta )\,\,=\,\,\frac{3+6i\,\,\sin \,\theta \,\,+\,\,2i\,\,\sin \,\theta \,-4\,{{\sin }^{2\,}}\,\theta }{1+4\,{{\sin }^{2}}\,\theta \,}\] \[=\,\,\,\frac{(3-4\,\,\sin {{\,}^{2}})+8i\,\sin \,\theta }{1+4\,{{\sin }^{2}}\,\theta \,}\] \[\therefore \,\,\,\,f\left( \theta \right)\] is purely imaginary \[\therefore \,\,\text{ }3-4\,\,si{{n}^{2}}\,\theta \,\,=\,\,0\] \[{{\sin }^{2}}\,\theta \,\,=\,\,\frac{3}{4}\] \[si{{n}^{2}}\,\theta \,\,\,=\,\,\pm \,\frac{\sqrt{3}}{4}\] \[\therefore \,\,\theta \,\,=\,\,\frac{-\pi }{3},\,\,\frac{\pi }{3},\,\,\frac{2\pi }{3}\]
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