<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] 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 $\sin x$.
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