Question

find the real value of x such that ( 3+2isinx)/(1−2isinx) is purely imaginary?please answer anyone?

Answer 1

Please see the attachment

Answer 2

The value of $x$ is $n\pi$

Step-by-step explanation:

Given,

$\frac{3+2.i.sinx}{1-2.i.sinx}$ is purely imaginary then find the value of $x$

∵ $\frac{3+2.i.sinx}{1-2.i.sinx}$ is purely imaginary.

So, we take imaginary part is $0$ after solving above equation.

∴ $\frac{3+2.i.sinx}{1-2.i.sinx}$

$=\frac{(3+2.i.sinx)(1+2.i.sinx)}{(1-2.i.sinx)(1+2.i.sinx)}$

$=\frac{3+6.i.sinx+2.i.sinx+4.i^{2}.sin^{2}x}{1-4.i^{2}.sin^{2}x}$

$=\frac{3+8.i.sinx+4.i^{2}.sin^{2}x}{1-4.i^{2}.sin^{2}x}$

Plug $i^{2}=-1$,

$=\frac{3+8.i.sinx-4sin^{2}x}{1+4sin^{2}x}$

$=\frac{3-4sin^{2}x }{1+4sin^{2}x}+i.\frac{8.sinx}{1+4sin^{2}x}$

So, Imaginary parts of is equal to $0$

∴ $\frac{8.sinx}{1+4.sin^{2}x} =0$

⇒$8.sinx=0$

⇒$sinx=0$

⇒$sinx=sin0$

∴ $x=n\pi \pm0$

⇒ $x=n\pi$ where $n\epsilon \mathbb{Z}$

So, The value of $x$ is $n\pi$