If $0<x<\pi$, and $\cos x+\sin x=1/2$, then $\tan x$ is

Clearly $x\ne \pi /2$ so $\cos x\ne 0$ and we have $2(1+\tan x)=\sec x$.

Squaring,

$\begin{array}{l}4(1+\tan x{)}^2=1+{\tan }^{2}x\\ \Rightarrow 3{\tan }^{2}x+8\tan x+3=0\\ \Rightarrow \tan x=\frac{ -8\pm \sqrt{64-36} }{6}=\frac{-4\pm \sqrt{7}}{3}\end{array}$

(Take the negative sign)