Find k that satisfies $\cos{(\omega fx)}=\cos{(\omega k x)}$

Question

For which frequency k>f is

$\cos (\omega fx_1)= \cos (\omega kx_1)$

only at point $x_1$.

Like k = 9

$\cos (\omega*1*0.1)= \cos (\omega * 9 * 0.1) \approx 0.809$

I need at general solution.

Answer 1

In general we have that:

$$\cos (A)= \cos (B) \iff A=B+2n\pi \;\lor \; A=-B +2n\pi \;, n\in \mathbb Z$$

and then

$$\cos (\omega fx_1)= \cos (\omega kx_1) \iff \omega fx_1=\omega kx_1+2n\pi \;\lor \; fx_1=-\omega kx_1 +2n\pi$$

Answer 2

Use that $$\cos(x)-\cos(y)=-2 \sin \left(\frac{x}{2}-\frac{y}{2}\right) \sin \left(\frac{x}{2}+\frac{y}{2}\right)$$ You will get $$-2\sin(\frac{wfx-wkx}{2})\sin(\frac{wfx+wkx}{2})=0$$ Not that $$\sin(x)=0$$ if $x=2k\pi$