If the function given by
\(
f(x)= \begin{cases}-2 \sin x & -\pi \leq x < -(\pi / 2) \ a \sin x+b \end{cases}
\) \(-(\pi / 2) < x < (\pi / 2) \ \cos x (\pi / 2) \leq x \leq \pi\)
is continuous in \([-\pi, \pi]\), then the value of \((3 a+2 b)^3\) is

\(
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x)=\lim _{x \rightarrow \frac{\pi^{-}}{2}}-2 \sin x=-2 \sin \left(-\frac{\pi}{2}\right)=2 \\
& \lim _{x \rightarrow \frac{-\pi^{+}}{2}} f(x)=\lim _{x \rightarrow \frac{-\pi^{+}}{2}} a \sin x+b=a \sin \left(\frac{-\pi}{2}\right)+b=-a
\end{aligned}
\)
Since \(f(x)\) is continuous at \(x=\frac{-\pi}{2}\), we write
\(
\begin{aligned}
& 2=-a+b \\
& \lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x) \lim _{x \rightarrow \frac{\pi^{-}}{2}} a \sin x+b=a \sin \left(\frac{\pi}{2}\right)+b=a+b \\
& \lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x)=\lim _{x \rightarrow \frac{\pi^{+}}{2}} \cos x=\cos \left(\frac{\pi}{2}\right)=0
\end{aligned}
\)
Since \(f(x)\) is continuous at \(x=\frac{\pi}{2}\), we write
\(
\mathrm{a}+\mathrm{b}=0
\)
From (1) and (2), we get \(\mathrm{a}=-1, \mathrm{~b}=1\)
\(
\therefore(3 a+2 b)^3=(-3+2)^3=-1
\)