trig

By sum-to-product formula:

$\sin \dfrac x2 + \sin \dfrac x3 = 2 \sin \dfrac{5x}{12} \cos\dfrac{x}{12}$

The function $g(x) = \sin \dfrac{5x}{12}$ has period $\dfrac{24\pi}5$, which means $g\left(\dfrac{24n\pi}5 + x\right) = g(x), n\in \mathbb Z$

The function $h(x) = \cos \dfrac{x}{12}$ has period 24 pi, which means $h(24n\pi + x) = h(x), n\in \mathbb Z$

Therefore $f(24\pi + x) = 2g(24\pi + x) h(24\pi + x) = 2g\left(\dfrac{24\pi}5 \cdot 5 + x\right)h(24\pi \cdot 1 + x) = 2g(x) h(x) = f(x)$

Therefore the period is $24\pi$.

No smaller T satisfies $f(T + x) = f(x)$