Nếu \(\cos x + \sin x = \frac{1}{2}\) và \({0^0} < x < {180^0}\) thì \(\tan x{\rm{ = }} - \frac{{p + \sqrt q }}{3}\) với c�

\(\begin{array}{l}
{\left( {\sin \alpha  + \cos \alpha } \right)^2} = {\sin ^2}\alpha  + 2\sin \alpha .\cos \alpha  + {\cos ^2}\alpha  = 1 + 2\sin \alpha .\cos \alpha \\
 \Rightarrow \sin \alpha .\cos \alpha  = \frac{{{{\left( {\sin \alpha  + \cos \alpha } \right)}^2} - 1}}{2} =  - \frac{3}{8}\\
 \Rightarrow \left\{ \begin{array}{l}
\sin \alpha  = \frac{{1 + \sqrt 7 }}{4}\\
\cos \alpha  = \frac{{1 - \sqrt 7 }}{4}
\end{array} \right. \Rightarrow \tan \alpha  =  - \frac{{4 + \sqrt 7 }}{3}
\end{array}\)

Suy ra (p;q)=(4;7)