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

$\cos x+\sin x=1/2$ Squaring both sides:${\cos }^{2}x+{\sin }^{2}x+2\sin x\cos x=1/4$$1+\sin 2x=1/4⟹\sin 2x=-3/4$ Since $\sin 2x$ is negative and $0<x<\pi$, $x$ must be in the second quadrant (obtuse).Using $\sin 2x=\frac{2\tan x}{1+{\tan }^{2}x}$:$-3/4=\frac{2\tan x}{1+{\tan }^{2}x}⟹-3-3{\tan }^{2}x=8\tan x$$3{\tan }^{2}x+8\tan x+3=0$$\tan x=\frac{-8\pm \sqrt{64-36}}{6}=\frac{-8\pm \sqrt{28}}{6}=\frac{-8\pm 2\sqrt{7}}{6}=\frac{-4\pm \sqrt{7}}{3}$ Since $x$ is obtuse, $\tan x$ is negative.Also, $\cos x+\sin x=1/2>0$, which for second quadrant requires $|\sin x|>|\cos x|$, meaning $x$ is closer to $\pi /2$ than $\pi$.Thus $\tan x=\frac{-4-\sqrt{7}}{3}$.