Explanation
We have,
$$\dfrac{2+3i\sin \theta }{1-2i\sin \theta }$$
On rationalize and we get,
$$ \dfrac{2+3i\sin \theta }{1-2i\sin \theta }\times \dfrac{1+2i\sin \theta }{1+2i\sin \theta } $$
$$ \Rightarrow \dfrac{2+3i\sin \theta +4i\sin \theta +6{{i}^{2}}{{\sin }^{2}}\theta }{{{1}^{2}}-4{{i}^{2}}{{\sin }^{2}}\theta } $$
$$ \Rightarrow \dfrac{2-{{\sin }^{2}}\theta +7i\sin \theta }{1+4{{\sin }^{2}}\theta } $$
$$ \Rightarrow \dfrac{1+1-{{\sin }^{2}}\theta +7i\sin \theta }{1+4{{\sin }^{2}}\theta } $$
$$ \Rightarrow \dfrac{1+{{\cos }^{2}}\theta +7i\sin \theta }{1+4{{\sin }^{2}}\theta } $$
$$ \Rightarrow \dfrac{1+{{\cos }^{2}}\theta }{1+4{{\sin }^{2}}\theta }+\dfrac{7i\sin \theta }{1+4{{\sin }^{2}}\theta } $$
$$ \Rightarrow \dfrac{1+{{\cos }^{2}}\theta }{1+4{{\sin }^{2}}\theta }+i\dfrac{7\sin \theta }{1+4{{\sin }^{2}}\theta } $$
Now,
Then purely imaginary is
Real part of equal to zero.
$$ \dfrac{1+{{\cos }^{2}}\theta }{1+4{{\sin }^{2}}\theta }=0 $$
$$ \Rightarrow 1+{{\cos }^{2}}\theta =0 $$
$$ \Rightarrow {{\cos }^{2}}\theta =-1 $$
$$ \Rightarrow \cos \theta =\sqrt{-1} $$
$$ \Rightarrow \theta ={{\cos }^{-1}}\sqrt{-1} $$
Hence, this is the answer.
$${\textbf{Step -1: Considering option A.}}$$
$${x^2} - 4x + 3\sqrt 2 = 0$$
$$\therefore D = {\left( { - 4} \right)^2} - 4\left( 1 \right)\left( {3\sqrt 2 } \right)$$
$$\therefore D = 16 - 12\sqrt 2 < 0$$
$${\text{Therefore, it has no real roots}}{\text{.}}$$
$${\textbf{Step -2: Considering option B.}}$$
$${x^2} + 4x - 3\sqrt 2 = 0$$
$$\therefore D = {\left( 4 \right)^2} - 4\left( 1 \right)\left( { - 3\sqrt 2 } \right)$$
$$\therefore D = 16 + 12\sqrt 2 > 0$$
$${\text{Therefore, it has real roots}}{\text{.}}$$
$${\textbf{Step -3: Considering option C.}}$$
$${x^2} - 4x - 3\sqrt 2 = 0$$
$$\therefore D = {\left( { - 4} \right)^2} - 4\left( 1 \right)\left( { - 3\sqrt 2 } \right)$$
$${\textbf{Step -4: Considering option D.}}$$
$$ 3{x^2} + 4\sqrt 3 x + 4 = 0$$
$$\therefore D = {\left( {4\sqrt 3 } \right)^2} - 4\left( 3 \right)\left( 4 \right)$$
$$\therefore D = 48 - 48$$
$$\therefore D = 0$$
$${\text{Therefore, it has real and equal roots}}{\text{.}}$$
$${\textbf{Hence, option A. is correct answer.}}$$
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