1. We are given

$$0<x<\pi, \qquad \sin x+\cos x=\frac12.$$

We need to find $\tan x$.

2. Use the identity

$$(\sin x+\cos x)^2=\sin^2 x+\cos^2 x+2\sin x\cos x=1+2\sin x\cos x.$$

Substitute the given value:

$$\left(\frac12\right)^2=1+2\sin x\cos x.$$

So,

$$\frac14=1+2\sin x\cos x$$ $$2\sin x\cos x=-\frac34$$ $$\sin x\cos x=-\frac38.$$

3. Since $0<x<\pi$, we have $\sin x>0$. Because $\sin x\cos x<0$, it follows that $\cos x<0$. Hence $x$ lies in the second quadrant, so

$$\tan x<0.$$

Therefore the answer must be negative. Among the options, only option C is negative.

4. Let

$$t=\tan x=\frac{\sin x}{\cos x}.$$

Now,

$$\sin x+\cos x=\frac12.$$

Write in terms of $t$:

$$\sin x=t\cos x.$$

Thus,

$$t\cos x+\cos x=\frac12$$ $$\cos x(1+t)=\frac12.$$

Squaring,

$$\cos^2 x(1+t)^2=\frac14.$$

But

$$\cos^2 x=\frac{1}{1+t^2}.$$

So,

$$\frac{(1+t)^2}{1+t^2}=\frac14.$$

Cross-multiply:

$$4(1+2t+t^2)=1+t^2.$$ $$4+8t+4t^2=1+t^2$$ $$3t^2+8t+3=0.$$

Solve the quadratic:

$$t=\frac{-8\pm\sqrt{64-36}}{6}=\frac{-8\pm\sqrt{28}}{6}=\frac{-8\pm 2\sqrt7}{6}=\frac{-4\pm \sqrt7}{3}.$$

So,

$$t=\frac{-4+\sqrt7}{3}\quad \text{or} \quad t=\frac{-4-\sqrt7}{3}.$$

5. Now use the quadrant condition. Since $x$ is in quadrant II, $\tan x<0$, and both are negative. We must check which one satisfies $\sin x+\cos x=\frac12$.

Using

$$\cos x=\frac{1}{\sqrt{1+t^2}},\quad \sin x=\frac{t}{\sqrt{1+t^2}}$$

for quadrant I only is not suitable directly here because in quadrant II, $\cos x<0$ and $\sin x>0$.

Instead, from

$$\cos x(1+t)=\frac12,$$

we need the right-hand side positive. Since in quadrant II, $\cos x<0$, we must have

$$1+t<0 \quad \Rightarrow \quad t<-1.$$

Now check the two roots:

$$\frac{-4+\sqrt7}{3}>-1$$

since $\sqrt7\approx 2.646$, giving about $-0.451$. This does not satisfy $t<-1$.

The other root:

$$\frac{-4-\sqrt7}{3}\approx -2.215<-1,$$

which is valid.

Hence,

$$\tan x=-\frac{4+\sqrt7}{3}.$$

6. Therefore the correct option is

$$\boxed{\text{C }\; -\frac{4+\sqrt7}{3}}.$$

7. Comparison with stored answer: Stored correct answer = C. This matches our derived answer.