$\text{ The period of the function }\left|{\sin }^3\frac{x}{2}\right|+\left|{\cos }^5\frac{x}{5}\right|\text{ is }$

$f\left(x\right)= \left|{\sin }^3\frac{x}{2}\right|+\left|{\cos }^5\frac{x}{5}\right|$

$the period of {\sin }^{3}x is 2\mathrm{\pi } \left(\mathrm{since} {\sin }^3\left(2\mathrm{\pi }+\mathrm{x}\right)={\sin }^3\mathrm{x}\right)$

$\Rightarrow the period of {\sin }^3\frac{x}{2}= \frac{2\mathrm{\pi }}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}} =4\mathrm{\pi } \left( \mathrm{if} \mathrm{f}\left(\mathrm{x} \right) \mathrm{period} \mathrm{is} \mathrm{p} \mathrm{the}n \mathrm{f}\left(\mathrm{ax}+\mathrm{b} \right) \mathrm{period} \mathrm{is} \frac{\mathrm{p}}{\left|\mathrm{a}\right|} \right)$

$\Rightarrow so the period of \left|{\sin }^3\frac{x}{2} \right| is 2\mathrm{\pi }$

$the period of {\cos }^{5}x is 2\mathrm{\pi }$

$so the period of {\cos }^5\frac{x}{5} is \frac{2\mathrm{\pi }}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}}= 10\mathrm{\pi }$

$so the period of \left|{\cos }^5\frac{x}{2}\right| is =5\mathrm{\pi }$

$so the period of f\left(x\right) = LCM of\left\{2\mathrm{\pi },5\mathrm{\pi }\right\} =10\mathrm{\pi }$