$\cos x + \sin x = {1 \over 2}$ $\Rightarrow {\left( {\cos x + {\mathop{\rm sinx}\nolimits} } \right)^2} = {1 \over 4}$ $\Rightarrow {\cos ^2}x + {\sin ^2}x + 2\cos x\sin x = {1 \over 4}$ $\left[ \because {{{\cos }^2}x + {{\sin }^2}x = 1\, \,and \,\,2\cos x\sin x = \sin 2x} \right]$ $\Rightarrow 1 + \sin 2x = {1 \over 4}$ $\Rightarrow \sin 2x = - {3 \over 4},$ so $x$ is obtuse and ${{2\tan x} \over {1 + {{\tan }^2}x}} = - {3 \over 4}$ $\Rightarrow 3{\tan ^2}x + 8\tan x + 3 = 0$ $\therefore$ $\tan x = {{ - 8 \pm \sqrt {64 - 36} } \over 6}$ $= {{ - 4 \pm \sqrt 7 } \over 3}$ as $\tan x < 0\,$ $\therefore$ $\tan x = {{ - 4 - \sqrt 7 } \over 3}$