Find the value of x which satisfies the equation $x=1-x+x^2-x^3+x^4-x^5+\dots$

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=1-x+x^2-x^3+x^4-x^5+\dots are in G.p\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=\frac{1}{(1+x)}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x+x^2-1=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{x}^2+x-1=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=\frac{-1\pm \sqrt{1+4}}{2}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left.x=\frac{-1+\sqrt{5}}{2}(\because -1<x<1)\right)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=\frac{\sqrt{5}-1}{2}=2\left(\frac{\sqrt{5}-1}{4}\right)=2\sin \left({18}^{\circ }\right)\end{array}$

Hence the solution is x = 2 sin (18°).