I am trying to solve t for the next equation:
$$A\sin(\omega_{1}t)+B\sin(\omega_{2}t)=0$$
Where:
$A > B$
$\omega_{1}$ AND $\omega_{2}$ are constants
Reading in wikipedia about trigonometric identities and following this post: Identity for a weighted sum of sines / sines with different amplitudes
I tried:
$$A\sin(\omega_{1}t)+B\sin(\omega_{2}t)=C\sin(\omega_{1}t+\varkappa)=0$$
Where:
$$C^2=A^2+B^2+2AB\cos(\omega_{2}t-\omega_{1}t)$$
And
$$\varkappa=\arcsin\frac{B\sin(\omega_{2}t-\omega_{1}t)}{C}$$
if $C\sin(\omega_{1}t+\varkappa)=0$, means that $C$ and/or $sin(\omega_{1}t+\varkappa)$ are equal to $0$, so I can get a partial solution by making $C=0$:
$$t=\frac{\arccos-\frac{A^2+B^2}{2AB}}{\omega_{2}-\omega_{1}}$$
But for the expression $sin(\omega_{1}t+\varkappa)=0$ I have not been able to solve it.
Someone can help me with this please?
Thanks