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 \(f(T + x) = f(x)\)