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.
 A: 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.  
A: Using the hint,
$$\Re(z+z^2-z^3)=1,$$ or
$$z+z^2-z^3=1+iw.$$
This can be factored as
$$-(z+1)(z-1)^2=iw$$ but I see no easy way to exploit it.
Direct resolution of the cubic equation looks terrible.
A: Hint:
$$\cos x+\cos2x=2\cos\dfrac{3x}2\cos\dfrac x2$$
$$1+\cos3x=2\cos^2\dfrac{3x}2$$
A: using the Addition formulas we get this here
$$4\,\cos \left( x \right) +2\, \left( \cos \left( x \right)  \right) ^{
2}-2-4\, \left( \cos \left( x \right)  \right) ^{3}
=0$$
A: Another idea ,maybe 
$$cos(x)+cos(2x)-cos(3x)=1\\
cos(x)-cos(3x)=1-\cos(2x)\\\cos
(2x-x)-\cos(2x+x)=1-\cos(2x)\\\cos(2x)\cos(x)+\sin(2x)\sin(x)-(\cos(2x)\cos(x)-\sin(2x)\sin(x))=1-\cos(2x)\\2\sin(2x)\sin(x)=1-\cos(2x)\\
2\sin(2x)\sin(x)=2\sin^2(\frac{2x}{2})\\
2\sin(2x)\sin(x)=2\sin^2(x)\\
\sin(2x)\sin(x)-\sin^2(x)=0\\\sin(x)(\sin(2x)-\sin(x))=0\\ \sin(x)=0 \to\\x=k\pi\\\sin(2x)=\sin(x) \to\\ 2x=x+2k\pi,2x=\pi-x+2k\pi$$
A: Hint #1: Use the sum to product formula
$$\cos A + \cos B = 2 \sin \frac {A+B}{2} \sin \frac {A-B}{2}$$
Hint #2: With Hint #1, write $\cos 2x$ in terms of $\sin x$ only...what do you notice?
Hint #3: With the results from Hint #2, factor the resulting equation.
