Square \(\cos x + \sin x = \frac{1}{2}\): \(1 + 2\sin x\cos x = \frac{1}{4}\), so \(\sin x\cos x = -\frac{3}{8}\).

Since \(\sin x\cos x = \frac{\tan x}{1 + \tan^2 x}\), let \(t = \tan x\):

\(\frac{t}{1+t^2} = -\frac{3}{8} \Rightarrow 3t^2 + 8t + 3 = 0 \Rightarrow t = \frac{-4 \pm \sqrt{7}}{3}\)

Which one? \(0 < x < \pi\) gives \(\sin x > 0\), and \(\sin x\cos x < 0\) gives \(\cos x < 0\). Since \(\sin x + \cos x > 0\), we have \(|\sin x| > |\cos x|\), so \(\tan x < -1\).

That means \(\tan x = \frac{-4 - \sqrt{7}}{3}\) (about −2.22).

Answer: (A)