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.

There are already some good solutions, but the following solution does use the OP's substitution.
I assume that $x$ is real, then
$$z=\cos(x)+i \sin(x), z^*=\cos(x)-i \sin(x)=z^{-1},$$ so the equation is equivalent to the following:
$$z+z^{-1}+z^2+z^{-2}-z^3-z^{-3}=2.$$
Let us use the following substitution: $q=z+z^{-1}$, so $q^2=z^2+2+z^{-2}$ and $q^3=z^3+3 z+3 z^{-1}+z^{-3}$, so the equation is equivalent to the following:
$$q+q^2-2-q^3+3q=2$$ or $$q^3-q^2-4 q+4=0.$$
The roots of this cubic equation are integer, so it is easy to find them.  
