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)!
 A: Hint: Can you write $\cos 2x$ and $\cos 3x$ in terms of $\cos x$?
A: To solve an "equals zero" equation involving polynomials or trigonometric functions, it's generally prudent to keep the $0$ on one side and only work with the other side. Here you could use the double and triple angle formulas (which can be deduced from the addition formulas if need be) given by
$$\cos2x=2\cos^2x-1,\qquad \cos3x=4\cos^3x-3\cos x.$$
Substitute these expressions in and your equation will read $P(\cos x)=0$ for a polynomial $P$. Solve for the roots of the polynomial and then take inverse cosines where allowed. Alternatively, if you're at all familiar with complex analysis you could forego the use of angle formulas and go right to a polynomial by using $\cos\theta=(e^{i\theta}+e^{-i\theta})/2$ and then solving for $e^{i\theta}$ and then $\theta$.
A: 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....
