If $\frac{\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)+i\tan x}{1+2i\sin \left(\frac{x}{2}\right)}$ is real then $x=$

$\frac{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)+i\tan x}{1+i\left(2\sin \frac{x}{2}\right)}$ is real

$\Rightarrow 2\sin \frac{x}{2}\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)-\tan x=0$

$\Rightarrow 2{\sin }^2\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}-\tan x=0$

$\Rightarrow 2{\sin }^2\frac{x}{2}+\sin x-\frac{\sin x}{\cos x}=0$

$\Rightarrow 1-\cos x+\sin x-\frac{\sin x}{\cos x}=0$

$\Rightarrow \cos x+{\cos }^{2}x+\sin x\cos x-\sin x=0$

 Where $\cos x\ne 0$

$\Rightarrow \cos x-\sin x+\sin x.\cos x-{\cos }^{2}x=0$

$\cos x\ne 0$

$\Rightarrow \left(\cos x-\sin x\right)-\cos x\left(\cos x-\sin x\right)=0$

Where $\cos x\ne 0$

$\Rightarrow \left(\cos x-\sin x\right)\left(1-\cos x\right)=0$

$\Rightarrow \cos x=\sin x$ (or) $\cos x=1$

$\Rightarrow x=n\pi +\frac{\pi }{4}$ (or) $x=2n\pi$

$\therefore$  solution is $x=\left(n\pi +\frac{\pi }{4}\right)\cup \left(2n\pi \right)$