Solving $\cos x + \cos 2x - \cos 3x = 1$ with the substitution $z = \cos x + i \sin x$ I need to solve 
$$\cos x+\cos 2x-\cos 3x=1$$
using the substitution$$z= \cos x + i \sin x $$
I fiddled around with the first equation using the double angle formula and addition formula to get 
$$\cos^2 x+4 \sin^2x\cos x-\sin^2 x=1$$ 
which gets me pretty close to something into which I can substitute $z$, because $$z^2= \cos^2 x-\sin^2 x+2i\sin x\cos x$$ 
I have no idea where to go from there.

Another idea ,maybe 
$$cos(x)+cos(2x)-cos(3x)=1\\
cos(x)-cos(3x)=1-\cos(2x)\\\cos
(2x-x)-\cos(2x+x)=1-\cos(2x)\\\cos(2x)\cos(x)+\sin(2x)\sin(x)-(\cos(2x)\cos(x)-\sin(2x)\sin(x))=1-\cos(2x)\\2\sin(2x)\sin(x)=1-\cos(2x)\\
2\sin(2x)\sin(x)=2\sin^2(\frac{2x}{2})\\
2\sin(2x)\sin(x)=2\sin^2(x)\\
\sin(2x)\sin(x)-\sin^2(x)=0\\\sin(x)(\sin(2x)-\sin(x))=0\\ \sin(x)=0 \to\\x=k\pi\\\sin(2x)=\sin(x) \to\\ 2x=x+2k\pi,2x=\pi-x+2k\pi$$
