Find $\tan x $ if $\sin x+\cos x=\frac12$ It is given that $0 < x < 180^\circ$ and $\sin x+\cos x=\frac12$, Find $\tan x $. 
I tried all identities I know but I have no idea how to proceed.  Any help would be appreciated. 

With no quadratic steps, only using trig identities:
Multiplying through by $\frac1{\sqrt{2}}$:
$$\begin{align}
\sin(x)\frac1{\sqrt{2}}+\cos(x)\frac1{\sqrt{2}}&=\frac{1}{\sqrt{8}}\\
\sin(x)\cos(\pi/4)+\cos(x)\sin(\pi/4)&=\frac{1}{\sqrt{8}}\\
\sin(x+\pi/4)&=\frac{1}{\sqrt{8}}
\end{align}$$
Given that $x$ is in $[0,\pi]$, $x$ must be in $(\pi/2,3\pi/4)$ because 


*

*if $x$ is in $[0,\pi/2]$ then both $\sin(x)$ and $\cos(x)$ are nonnegative, and $\max(\sin(x),\cos(x))\geq\frac{1}{\sqrt{2}}>\frac12$

*if $x$ is in $[3\pi/4,\pi]$ then $\cos(x)$ is negative and outweighs $\sin(x)$


So $x+\pi/4$ is in $(3\pi/4,\pi)$, and: $$\begin{align}
\sin(x+\pi/4)&=\frac{1}{\sqrt{8}}\\
x+\pi/4&=\pi-\arcsin\left(\frac{1}{\sqrt{8}}\right)\\
x&=\pi-\arcsin\left(\frac{1}{\sqrt{8}}\right)-\pi/4\\
\tan(x)&=\tan\left(\pi-\arcsin\left(\frac{1}{\sqrt{8}}\right)-\pi/4\right)\\
\tan(x)&=-\tan\left(\arcsin\left(\frac{1}{\sqrt{8}}\right)+\pi/4\right)\\
\tan(x)&=-\frac{\tan\left(\arcsin\left(\frac{1}{\sqrt{8}}\right)\right)+1}{1-\tan\left(\arcsin\left(\frac{1}{\sqrt{8}}\right)\right)}\\
\tan(x)&=-\frac{\frac{\frac{1}{\sqrt{8}}}{\sqrt{1-\frac{1}{8}}}+1}{1-\frac{\frac{1}{\sqrt{8}}}{\sqrt{1-\frac{1}{8}}}}\\
\tan(x)&=-\frac{\frac{1}{\sqrt{8}}+\sqrt{\frac78}}{\sqrt{\frac78}-\frac{1}{\sqrt{8}}}\\
\tan(x)&=\frac{1+\sqrt{7}}{1-\sqrt{7}}\\\end{align}$$
