cos x cos 2x cos 3x=1
The only way I can construct a solution is if
$\displaystyle cos(x) = 1$
and
$\displaystyle cos(2x) = 1$
and
$\displaystyle cos(3x) = 1$
or any two of these is equal to -1 and the remaining is equal to 1.
The solution to the first set of equations is, for example
$\displaystyle x = \frac{\pi}{2} + 2n \pi$
and
$\displaystyle x = \frac{\pi}{4} + k \pi$
and
$\displaystyle x = \frac{\pi}{6} + p \frac{2\pi}{3}$
respectively where n, k, and p are integers.
So:
$\displaystyle k = 2n + \frac{1}{4}$
and
$\displaystyle p = 3n + \frac{1}{2}$
Since k and p have to be integral there can be no solutions of this kind.
See what the other two solution options give you.
-Dan
Let $\displaystyle \mathrm{f}(x)=\cos{x}\cos{2x}\cos{3x}$. We need only consider $\displaystyle 0\leq x < \pi$ since $\displaystyle \mathrm{f}(x)$ has period $\displaystyle \pi$ (i.e. $\displaystyle \mathrm{f}(x+\pi)=\mathrm{f}(x)$).
For $\displaystyle x\in[0,\pi)$, $\displaystyle \cos{x}$ is equal to ±1 only when $\displaystyle x=0$. And indeed $\displaystyle \mathrm{f}(0)=1$. So that’s the only solution in that interval, and since f has period $\displaystyle \pi$, the complete solution set is $\displaystyle \{n\pi:n\in\mathbb{Z}\}$.