A good way to solve this trigonometric equation $$\sin x+\cos x=\frac{1}{2}$$
What is the value of $\tan x$? I tried using 
$$\sin2 x=\frac{2\tan x}{1+\tan^2x}$$ and $$\cos2 x=\frac{1-\tan^2x}{1+\tan^2x}$$ but we get a quadratic for $\tan\left(\frac{x}{2}\right)$ . So any better approach would be much appreciated. Thanks!

Given $$\sin x+\cos x=\frac{1}{2}$$
squaring both the sides, $$(\sin x+\cos x)^2=\left(\frac{1}{2}\right)^2$$
$$\sin^2 x+\cos^2 x+2\sin x\cos x=\frac{1}{4}$$
$$1+\sin 2x=\frac{1}{4}$$$$\sin 2x=-\frac{3}{4}$$
$$\frac{2\tan x}{1+\tan ^2x }=-\frac{3}{4}$$
$$3\tan^2 x+8\tan x+3=0$$
let $\tan x=t$ $$3t^2+8t+3=0$$
I hope you can take it from here.
