Solution:

Here $\Omega_1=5, \Omega_2=4$

$\begin{aligned}\\ & T_1=\frac{2 \pi}{\Omega_1}=\frac{2 \pi}{5}=\frac{2 \pi}{5} \\\\ & T_2=\frac{2 \pi}{\Omega_2}=\frac{2 \pi}{4}=\frac{\pi}{2} \\ \end{aligned} \\ $

$\frac{T_1}{T_2}=\frac{\frac{2 \pi}{5}}{\frac{\pi}{2}}=\frac{4}{5} \text { (It is rational number) }\\ $

Hence $x(t)$ is periodic,

$T=5 T_1=4 T_2=2 \pi\\ $

$ \therefore x(t) \text { is periodic with period …