Explanation

(b)
Since, $\frac{\left( \sin{\frac{x}{2} + \cos\frac{x}{2}} \right) - i\tan x}{1 + 2i\sin\frac{x}{2}} \in R$
$\Rightarrow \ \ \ \ \ \ \ \frac{\{\sin{\frac{x}{2} + \cos{\frac{x}{2} - i\tan x}}\}\left\{ 1 - 2i\sin\frac{x}{2} \right\}}{1 + 4\sin^{2}\frac{x}{2}} \in R$
It will be real, if imaginary part is zero
$\therefore\ \ - 2\sin{\frac{x}{2}\left\{ \sin{\frac{x}{2} + \cos\frac{x}{2}} \right\}} - \tan{x = 0}$
$\Rightarrow \ \ \ \ 2\sin{\frac{x}{2}\left\{ \sin{\frac{x}{2} + \cos\frac{x}{2}} \right\} + \frac{\sin x}{\cos x} = 0}$
$\Rightarrow \ \sin{\frac{x}{2}\left\lbrack \left\{ \sin{\frac{x}{2} + \cos\frac{x}{2}} \right\}\left\{ \cos^{2}{\frac{x}{2} - \sin^{2}\frac{x}{2}} \right\} + \cos\frac{x}{2} \right\rbrack} = 0$
$\therefore\ \sin{\frac{x}{2} = 0}$
$\Rightarrow \ \ \ \ x = 2n\pi$ …(i)
or $\left\{ \sin{\frac{x}{2} + \cos\frac{x}{2}} \right\}\left\{ \cos^{2}{\frac{x}{2} - \sin^{2}\frac{x}{2}} \right\} + \cos{\frac{x}{2} = 0}$
on dividing by $\cos^{3}\frac{x}{2}$
$\left( \tan{\frac{x}{2} + 1} \right)\left( 1 - \tan^{2}\frac{x}{2} \right) + \left( 1 + \tan^{2}\frac{x}{2} \right) = 0$
$\Rightarrow \ \ \ \tan^{3}{\frac{x}{2} - \tan{\frac{x}{2} - 2 = 0}}$
Let $\tan{\frac{x}{2} = t},$ then $f(t) = t^{3} - t - 2$,
Then $f(1) = - 2 < 0$ and $f(2) = 4 > 0$
Thus, $f(t)$ changes sign from negative to positive in (1, 2)
$\therefore$ Let $t = k$ be the root for which $f(k) = 0$ and $k \in (1,\ 2)$
$\therefore\ \ \ \ \ t = k$ or $\tan{\frac{x}{2} = k = \tan\alpha}$
Hence, $\frac{x}{2} = n\pi + \alpha$
$\Rightarrow \ \ \ \ \ \ \left\{ \begin{array}{r}
x = 2n\pi + 2\alpha\ \ \ \ \ \alpha = \tan^{- 1}k \\
or\ \ \ x = 2n\pi
\end{array} \right.\$ where $k \in (1,\ 2)$