<s>[INST] How to solve $\cos(x)\cos(2x)\cos(4x)=1/8$ I have to solve $\cos(x)\cos(2x)\cos(4x)=1/8$.
I can express it for $x$ only with $\cos(2x)=\cos^2(x)-\sin^2(x)$ and $\cos(4x)=\cos(2x+2x)$, but it only seems to become a really big expression and I have no clue how to proceed after... Any suggestions?
 [/INST] Let $z=e^{ix}$. Then your relation says $$\left(z+z^{-1}\right)\left(z^2+z^{-2}\right)\left(z^4+z^{-4}\right)=1$$ That is: 
$$
\begin{align}
z^7+z^5+z^3+z+z^{-1}+z^{-3}+z^{-5}+z^{-7}&=1\\
z^{-7}\frac{z^{16}-1}{z^2-1}&=1&\text{(for $z^2\neq1$)}\\
z^{16}-1&=z^9-z^7\\
z^{16}-z^9+z^7-1&=0\\
\left(z^7-1\right)\left(z^9+1\right)&=0
\end{align}$$ 
So $z$ is either a $7$th root of $1$ or a $9$th root of $-1$. Which means your $x$ is among $$\left\{\frac{2\pi}{7}, \frac{4\pi}{7},\frac{6\pi}{7},\frac{8\pi}{7},\frac{10\pi}{7},\frac{12\pi}{7},\frac{\pi}{9},\frac{3\pi}{9},\frac{5\pi}{9},\frac{7\pi}{9},\frac{11\pi}{9},\frac{13\pi}{9},\frac{15\pi}{9},\frac{17\pi}{9}\right\}$$ or translates by $2\pi k$. We've left out $0\pi/7$ and $9\pi/9$ since they cause $z^2-1$ to equal $0$, invalidating an earlier step here. You can directly check that these angles do not satisfy the original equation.
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