7.9 If $\sin x+\cos x=\frac{1}{5}$, and $0 \leqslant x<\pi$, then $\operatorname{tg} x$ is
(A) $-\frac{4}{3}$.
(B) $-\frac{3}{4}$.
(C) $\frac{3}{4}$.
(D) $\frac{4}{3}$.
(E) Cannot be determined completely from the given conditions.
(29th American High School Mathematics Examination, 1978)
[Solution] If $\sin x + \cos x = \frac{1}{5}$, then from $\sin^2 x + \cos^2 x = 1$ we have

that is,
$$
\begin{array}{l}
1 - \sin^2 x = \left(\frac{1}{5} - \sin x\right)^2, \\
25 \sin^2 x - 5 \sin x - 12 = 0.
\end{array}
$$

Similarly, $\cos x$ satisfies the equation $25 t^2 - 5 t - 12 = 0$, whose solutions are $\frac{4}{5}$ and $-\frac{3}{5}$.
Since $\sin x \geqslant 0$, we get $\sin x = \frac{4}{5}$;
and since $\cos x = \frac{1}{5} - \sin x$,
we get $\cos x = -\frac{3}{5}$. Therefore, $\operatorname{tg} x = -\frac{4}{3}$. Hence, the answer is $(A)$.