Find $ such that (3 + 2i sin $)/ (1- 2i sin$) is purely imaginary number ?Answer with solution please .

I am replacing $ with x.

For the expression to be purely imaginary,

(3 + 2i sinx )/ (1 - 2i sinx) = Bi for some real number B.

=> 3 + 2i sin x = 2B sin x + Bi

Comparing the real and imaginary parts of LHS and RHS, we get

3 = 2B sin x; ……………………………(i)

2sinx = B………………………………..(ii)

Using the value of B obtained from (ii) in (i),

3 = 4 (sin x) ^2

=> (sin x) ^2 = 3/4

=> sin x = +sqrt(3)/2 , -sqrt(3)/2

=> x = Pi/3 + 2nPi, 2Pi/3 + 2nPi, -Pi/3 + 2nPi, 4Pi/3 + 2nPi, 5Pi/3 + 2nPi

(Where n is any integer)

If the answers are asked in the range [0,2pi], the solutions are:

Pi/3, 2Pi/3, 4Pi/3, 5Pi/3