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!
 A: Where does the line $x + y = \frac{1}{2}$ intersect the unit circle?
A: 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.
A: HINT....Write the LHS as$$ \sqrt{2}\sin(x+\frac {\pi}{4})$$
You can find $x$ then find $\tan x$
Usually the best way to proceed when you have an equation of the form $$a\sin x+b\cos x=c$$ is to rewrite the LHS as $$R\cos(x+\alpha)$$ or $$R\sin(x+\alpha)$$. This is known as a compound angle transformation.
A: HINT:
$$\cos(x)+\sin(x)=\frac{1}{2}\Longleftrightarrow$$
$$\sqrt{2}\left(\frac{\cos(x)}{\sqrt{2}}+\frac{\sin(x)}{\sqrt{2}}\right)=\frac{1}{2}\Longleftrightarrow$$
$$\sqrt{2}\left(\sin\left(\frac{\pi}{4}\right)\cos(x)+\cos\left(\frac{\pi}{4}\right)\sin(x)\right)=\frac{1}{2}\Longleftrightarrow$$
$$\sqrt{2}\sin\left(\frac{\pi}{4}+x\right)=\frac{1}{2}\Longleftrightarrow$$
$$\sin\left(\frac{\pi}{4}+x\right)=\frac{1}{2\sqrt{2}}\Longleftrightarrow$$
$$\frac{\pi}{4}+x=\pi-\sin^{-1}\left(\frac{1}{2\sqrt{2}}\right)+2\pi n_1\Longleftrightarrow\space\space\vee\space\space\frac{\pi}{4}+x=\sin^{-1}\left(\frac{1}{2\sqrt{2}}\right)+2\pi n_2\Longleftrightarrow$$
$$x=2\left(\pi n_1+\tan^{-1}\left(\frac{1}{3}\left(2-\sqrt{7}\right)\right)\right)\space\space\vee\space\space x=2\left(\pi n_2+\tan^{-1}\left(\frac{1}{3}\left(2+\sqrt{7}\right)\right)\right)$$
With $n_1,n_2\in\mathbb{Z}$
