trying to solve $\sqrt{\cos(x)-2\cos(2x)}+\sqrt{2}\cos(2x)=0$ The equation is
$$\sqrt{\cos(x)-2\cos(2x)}+\sqrt{2}\cos(2x)=0$$
The system is
$$
\begin{cases}
\cos(x)-2\cos(2x)=2\cos^2(2x) \\
-\sqrt{2}\cos(2x)\ge 0 \iff \cos(2x)\le 0
\end{cases}
$$
The equation:
$$\cos(x)-2(2\cos^2(x)-1)=2(2\cos^2(x)-1)^2$$
$$\cos(x)-4\cos^2(x)+2=2(4\cos^4(x)-4\cos^2(x)+1)$$
$$\cos(x)-4\cos^2(x)+2=8\cos^4(x)-8\cos^2(x)+2$$
$$8\cos^4(x)-4\cos^2(x)-\cos(x)=0$$
But what to do next? 
If I factor out $\cos(x)$, I get a non-factorizable third-degree polynomial in parentheses
$$\cos(x)(8\cos^3(x)-4\cos(x)-1)=0$$

The problem with its solution (upside-down) from the textbook


Hint: You can factorise: $$8c^4-4c^2-c=c(2c+1)(4c^2-2c-1)$$
