If $\sin x+\cos x=1/5$ and $0\le x<\pi$, then $\tan x$ is
$\textbf{(A) }-\frac{4}{3}\qquad \textbf{(B) }-\frac{3}{4}\qquad \textbf{(C) }\frac{3}{4}\qquad \textbf{(D) }\frac{4}{3}\qquad\\ \textbf{(E) }\text{not completely determined by the given information}$
Squaring the equation, we get
\[\sin ^2 x + \cos ^2 x + 2 \sin x \cos x = \frac{1}{25} =\Rrightarrow 2\sin x \cos x = -\frac{24}{25} \Rrightarrow \sin x \cos x = -\frac{12}{25}\]
Recall that $(a-b)^2 = (a+b)^2 - 4ab$, so
\[(\sin x - \cos x)^2 = (\sin x + \cos x)^2 - 4 \sin x \cos x\]
\[(\sin x - \cos x)^2 = \frac{1}{25} - 4 (- \frac{12}{25})\]
\[(\sin x - \cos x) = \frac{7}{5}\]
We can now solve for the values of $\sin x$ and $\cos x$
\[\sin x + \cos x = \frac{1}{5}\]
\[\sin x - \cos x = \frac{7}{5}\]
\[2 \sin x = \frac{8}{5} \Rrightarrow \sin x = \frac{4}{5}, \cos x = -\frac{3}{5}\]
Since $\tan x = \frac{\sin x}{\cos x}$, we have
\[\tan x = \frac{\frac{4}{5}}{-\frac{3}{5}} = -\frac{4}{3}\]
$\boxed{A}$
~JustinLee2017