13. From elementary trigonometry we have $\sin 3 t=3 \sin t-4 \sin ^{3} t$. Hence, if we denote $y=\sin 20^{\circ}$, we have $\sqrt{3} / 2=\sin 60^{\circ}=3 y-4 y^{3}$. Obviously $00$ for $0 \leq x<1 / 2$. Now the desired inequality $\frac{20}{60}=\frac{1}{3}<\sin 20^{\circ}<\frac{21}{60}=\frac{7}{20}$ follows from $$f\left(\frac{1}{3}\right)<\frac{\sqrt{3}}{2}<f\left(\frac{7}{20}\right)$$ which is directly verified.