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?
 A: You can use the double angle identity
$$ \sin 2x = 2\sin x \cos x
$$
By multiplying $\sin x$,
$$\begin{align*}
\cos(x)\cos(2x)\cos(4x) &= \frac{1}{8} \\
\frac{1}{2} \sin (2x) \cos(2x)\cos(4x) &= \frac{1}{8}\sin x \\
\frac{1}{4} \sin (4x)\cos(4x) &= \frac{1}{8}\sin x \\
\frac{1}{8} \sin (8x) &= \frac{1}{8}\sin x \\
\sin (8x) &= \sin x
\end{align*}$$
The last equation will have $16$ roots in $[0,2\pi)$, but $0$ and $\pi$ do not solve the original equation since they are introduced by $\sin x$.
A: For $x\ne n\pi$ (it's true, because $ \cos{x}\cdot\cos{2x}\cdot\cos{4x}\,\Big|_{x=n\pi} \ne\frac{1}{8}$)
$$ \cos{x}\cdot\cos{2x}\cdot\cos{4x}=\frac{\sin{x}\cdot\cos{x}\cdot\cos{2x}\cdot\cos{4x}}{\sin{x}} = \\
=\frac{1}{2}\cdot \frac{\sin{2x}\cdot\cos{2x}\cdot\cos{4x}}{\sin{x}}=\frac{1}{4}\cdot \frac{\sin{4x}\cdot\cos{4x}}{\sin{x}}=\frac{1}{8}\cdot \frac{\sin{8x}}{\sin{x}},$$
therefore,
$$\sin{8x}=\sin{x}.$$
A: 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.
