A displacement is x(t) = 3 sin(4π t) + 2 cos(6π t), with x in cm and t in s…

Step 1. Each part. The first part has $\omega = 4\pi\ \text{rad/s}$, so $T_1 = \dfrac{2\pi}{4\pi\ \text{rad/s}} = 0.5\ \text{s}$. The second has $\omega = 6\pi\ \text{rad/s}$, so $T_2 = \dfrac{1}{3}\ \text{s}$.

Step 2. Both together. We need a time that holds a whole number of each. The multiples of $0.5\ \text{s}$ are $0.5\ \text{s}$, $1\ \text{s}$, $1.5\ \text{s}$ and so on. The multiples of $\tfrac13\ \text{s}$ are $\tfrac13\ \text{s}$, $\tfrac23\ \text{s}$, $1\ \text{s}$ and so on. The first one in both lists is $1\ \text{s}$.