If  $0⩽x<\mathrm{\pi }$  and $\cos x+\sin x$$=\frac{1}{2}$, then  $\tan x$ =

$\cos x+\sin x$$=\frac{1}{2}$

$\Rightarrow \frac{\cos x}{\sqrt{2}}+\frac{\sin x}{\sqrt{2}} \Rightarrow \cos \left(x-\frac{\mathrm{\pi }}{4}\right)=\frac{1}{2\sqrt{2}}$

$\Rightarrow \tan \left(x-\frac{\mathrm{\pi }}{4}\right)$$=\sqrt{7 }$$\Rightarrow \frac{\tan x-1}{\tan x+1}=\frac{\sqrt{7}}{1}$

By componendo & dividendo , we get

$\frac{\tan x}{1}$ $=\frac{\sqrt{7 }+1}{1-\sqrt{7}}$$=\frac{(\sqrt{7}+1)(1+\sqrt{7})}{-6}$

$\therefore \tan x=-\frac{1}{6}(8+2\sqrt{7})=-\frac{(4+\sqrt{7})}{3}$