<s>[INST] 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

 [/INST] HINT:
$$8\cos^3x-4\cos x-1=4\cos x(2\cos^2x-1)$$
$$\implies2(\cos3x+\cos x)=1\iff4\cos x\cos2x=1$$
If $\sin x=0,4\cos x\cos2x\ne1\implies\sin x\ne0$
$$4\cos x\cos2x=1\iff\sin x=2(2\sin x\cos x)\cos2x=2\sin2x\cos2x=\sin4x$$
$$\sin4x=\sin x\implies4x=n\pi+(-1)^nx$$ where $n$ is any integer
Also, $\cos2x\le0\implies2m\pi+\dfrac\pi2\le2x\le2m\pi+\dfrac{3\pi}2$
</s>