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Question

Find real $$\theta$$ such that $$\dfrac{3+2i\sin{\theta}}{1-2i\sin{\theta}}$$, is purely Real.


Solution

$$\Rightarrow\dfrac{{3 + 2i\sin \theta }}{{1 - 2i\sin \theta }}$$

$$\Rightarrow \dfrac{{\left( {3 + 2i\sin \theta } \right)\left( {1 + 2i\sin \theta } \right)}}{{\left( {1 - 2i\sin \theta } \right)\left( {1 + 2i\sin \theta } \right)}}$$

$$\Rightarrow \dfrac{{\left( {3 - 4{{\sin }^2}\theta } \right)\left( {6\sin \theta  + 2\sin \theta } \right)i}}{{\left( {1 - 4{{\sin }^2}\theta } \right)}}$$

For purely real,

$$\Rightarrow \dfrac{{\left( {6\sin \theta  + 2\sin \theta } \right)}}{{\left( {1 - 4{{\sin }^2}\theta } \right)}} = 0$$

$$\Rightarrow 8\sin {\theta}=0$$

$$\Rightarrow \sin {\theta}=0$$

$$\Rightarrow{\theta}=0,\pi, 2\pi $$

Mathematics

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