Trigonometric equality $\cos x + \cos 3x - 1 - \cos 2x = 0$ In my text book I have this equation:
\begin{equation}
\cos x + \cos 3x - 1 - \cos 2x = 0
\end{equation}
I tried to solve it for $x$, but I didn't succeed.
This is what I tried:
\begin{align}
\cos x + \cos 3x - 1 - \cos 2x &= 0 \\
2\cos 2x \cdot \cos x - 1 - \cos 2x &= 0 \\
\cos 2x \cdot (2\cos x - 1) &= 1
\end{align}
So I clearly didn't choose the right path, since this will only be useful if I become something like $a \cdot b = 0$.
All tips will be greatly appreciated.

Solution (Addition to the accepted answer):

The problem was in writing $\cos 3x$ in therms of $cos x$. Anon pointed out that it was equal to $4\cos^3 x - 3\cos x$ but I had to work may way thru it to actually prove that. So I write it down here, maybe it's of use to anybody else.
\begin{align}
\cos 3x &= \cos(2x + x)\\
&= \cos(2x)\cdot \cos x - \sin(2x)\sin x\\
&= (\cos^2 x - \sin^2 x) \cdot \cos x - 2\sin^2 x \cdot \cos^2 x\\
&= \cos^3 x - \sin^2x\cdot \cos x - 2\sin^2x\cdot \cos^2 x\\
&= \cos^3 x - (1 - \cos^2 x)\cos x - 2(1-\cos^2 x)\cos x\\
&= cos^3 x - \cos x + \cos^3 x - 2\cos x + 2\cos^3 x\\
&= 4\cos^3 x - 3\cos x
\end{align}
EDIT: Apparently you can write all of the formulas $\cos(n\cdot x)$ with $n \in \{1, 2, 3, …\}$ in terms of $\cos x$. Why didn't my teacher tell me that! I don't have time to proof it myself now, but I'll definitely adapt my answer tomorrow (or any time soon)!

You can also use the identity: 
$$\cos(A)+\cos(B) = 2 \cos(\frac{A+B}{2}) \cos (\frac{A-B}{2})$$
Thus
\begin{equation}
\begin{split}
\cos x + \cos 3x - 1 - \cos 2x &= \cos x + \cos 3x - \cos(0) - \cos 2x \
&= 2\cos(\frac{x+3x}{2})\cos(\frac{x-3x}{2})- 2\cos(\frac{0+2x}{2})\cos(\frac{0-2x}{2}) 
\end{split}
\end{equation}
Then $2 \cos(x)$ becomes a common factor, and you can use the corresponding formula for difference this time to factor the rest....
