Find all possible values of $x$ if $\frac{\sin\frac{x}{2}+\cos\frac{x}{2}-i\tan x}{1+2i\sin\frac{x}{2}}$ is real 
If the expression $\dfrac{\sin\frac{x}{2}+\cos\frac{x}{2}-i\tan x}{1+2i\sin\frac{x}{2}}$ is real, find the set of all possible values of $x$.

My Attempt
$$
-\tan x-2\sin^2\frac{x}{2}-2\sin\frac{x}{2}\cos\frac{x}{2}=0\\
\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{\cos x}+2\sin^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}=0\\
\sin\frac{x}{2}=0\text{ or }\frac{\cos\frac{x}{2}}{\cos x}+\sin\frac{x}{2}+\cos\frac{x}{2}=0\\
\frac{x}{2}=n\pi\text{ or }\cos\frac{x}{2}+\sin\frac{x}{2}.\cos x+\cos\frac{x}{2}.\cos x=0\\
\boxed{x=2n\pi}\text{ or _____________}
$$
My reference gives the solution $x=2n\pi$, but what about the other remaining expression ?

You are right: there are more solutions. Actually, there is exactly one more solution in $(0,2\pi)$. Note that\begin{align}-\tan x-2\sin^2\frac{x}{2}-2\sin\frac{x}{2}\cos\frac{x}{2}=0&\iff-\tan x-1+1-2\sin^2\left(\frac x2\right)-2\sin x=0\\&\iff-\tan x-1+\cos x-\sin x.\end{align}Now, let $c=\cos x$ and let $s=\sin x$. Then$$\left\{\begin{array}{l}c^2+s^2=1\\-\dfrac sc-1+c-s=0\end{array}\right.$$Solving this system leads you to a quartic. It has a trivial solution ($(c,s)=(1,0)$) and a single real non-trivial solution: $(c,s)\simeq(-0.396608,0.917988)$. This corresponds to taking $x\simeq1.97861$.
