The function \(f(x)=4 \sin ^3 x-6 \sin ^2 x\)
\(+12 \sin x+100\) is strictly

Given, \(f(x)=4 \sin ^3 x-6 \sin ^2 x+12 \sin x+100\)
\[
\begin{aligned}
f^{\prime}(x) & =12 \sin ^2 x \cdot \cos x-12 \sin x \cos x+12 \cos x \\
& =12 \cos x\left[\sin ^2 x-\sin x+1\right] \\
& =12 \cos x\left[\sin ^2 x+(1-\sin x)\right]
\end{aligned}
\]
We know, \(1-\sin x \geq 0\) and \(\sin ^2 x \geq 0\)
\[
\therefore \sin ^2 x+1-\sin x>0
\]
Therefore, \(f^{\prime}(x)>0\), when \(\cos x>0 \Rightarrow x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)
So, \(f(x)\) is increasing when \(x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)
and \(f^{\prime}(x) < 0\), when \(\cos x < 0 \Rightarrow x \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)\)
So, \(f(x)\) is decreasing when \(x \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)\)
Hence, \(f(x)\) is decreasing in \(\left(\frac{\pi}{2}, \pi\right)\).