If
\[
f(x)=
\begin{cases}
\dfrac{1-\sin^3 x}{3\cos^2 x}, & x\lt \dfrac{\pi}{2} \\[6pt]
\alpha, & x=\dfrac{\pi}{2} \\[6pt]
\dfrac{\beta(1-\sin x)}{(\pi-2x)^2}, & x\gt \dfrac{\pi}{2}
\end{cases}
\]
is continuous at \(x=\dfrac{\pi}{2}\), then \(\alpha\beta=\)