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)!