# Math Help - solve

1. ## solve

cos x cos 2x cos 3x=1

2. Originally Posted by perash
cos x cos 2x cos 3x=1
The only way I can construct a solution is if
$cos(x) = 1$
and
$cos(2x) = 1$
and
$cos(3x) = 1$

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
$x = \frac{\pi}{2} + 2n \pi$
and
$x = \frac{\pi}{4} + k \pi$
and
$x = \frac{\pi}{6} + p \frac{2\pi}{3}$
respectively where n, k, and p are integers.

So:
$k = 2n + \frac{1}{4}$
and
$p = 3n + \frac{1}{2}$

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

3. Let $\mathrm{f}(x)=\cos{x}\cos{2x}\cos{3x}$. We need only consider $0\leq x < \pi$ since $\mathrm{f}(x)$ has period $\pi$ (i.e. $\mathrm{f}(x+\pi)=\mathrm{f}(x)$).

For $x\in[0,\pi)$, $\cos{x}$ is equal to ±1 only when $x=0$. And indeed $\mathrm{f}(0)=1$. So that’s the only solution in that interval, and since f has period $\pi$, the complete solution set is $\{n\pi:n\in\mathbb{Z}\}$.