Slight problem with solving a trigonometric equation. I had to prove the identity $4\cos^3x-3\cos x=\cos 3x$ and then use it to solve the equation $(4\cos^2x-3)(4\cos^23x-3)(4\cos^29x-3)=1$.
After proving the identity I proceeded to simplify the equation, obtaining
\begin{align*}
(4\cos^2x-3)(4\cos^23x-3)(4\cos^29x-3)&=1\\
(4\cos^3x-3\cos x)(4\cos^23x-3)(4\cos^29x-3)&=\cos x\\
\cos3x(4\cos^23x-3)(4\cos^29x-3)&=\cos x\\
(4\cos^33x-3\cos3x)(4\cos^29x-3)&=\cos x\\
\cos9x(4\cos^29x-3)&=\cos x\\
4\cos^39x-3\cos9x&=\cos x\\
\cos27x&=\cos x.
\end{align*}
The only thing is that when I plotted $y=(4\cos^2x-3)(4\cos^23x-3)(4\cos^29x-3)-1$ (the original equation which appears in red) and $y=\cos27x-\cos x$ (this is in black) the following graph emerged

Notice that after $1.5$ and $-1.5$ there is a root that is not shared by both curves. What do I have wrong?
 A: By multiplying by $\cos(x)$ you created additional solutions precisely at the points where $\cos(x)=0$. This is because at these points you just multiplied your original equation by zero which makes any equation true. Consider for example the equation $x = 1$ which has precisely one solution. Contrast this with the equation we get after multiplying with $\cos(x)$:
$$x\cos(x) = \cos(x)$$
which has infinitely many solutions, namely, for all $x$ such that $\cos(x)$ the equation is solved.
UPDATE:
Your method of finding solution is correct except that all solutions such that $\cos(x)=0$ are not necessarily solutions at all. This is because of what I said before. These solutions are $x = \frac{\pi}{2}+ n \pi$. You can manually check whether these $x$ values give solutions though by just working out the expression
$$(4\cos^2x-3)(4\cos^23x-3)(4\cos^29x-3)$$
and check if it equals $1$. You will find that this is not the case, as all cosine terms disappear and you are left with $(-3)(-3)(-3) \neq 1$. Thus you must remove from your set of solutions all $x$ such that $x = \frac{\pi}{2}+ n \pi$.
