The function f(x) = 4 sin³ x – 6 sin²x + 12 sin x + 100 is strictly

The correct option is: decreasing in [−π/2,π/2]

To determine the intervals of increase or decrease for the function $f\left(x\right)=4{\sin }^{3}x-6{\sin }^{2}x+12\sin x+100$, we need to find its first derivative, $f\text{'}\left(x\right)$, with respect to $x$.

Let $t=\sin x$. Since the options discuss intervals of $x$, let us first differentiate $f\left(x\right)$ directly using the chain rule:
$$f\text{'}\left(x\right)=\frac{d}{dx}[4{\sin }^{3}x-6{\sin }^{2}x+12\sin x+100]$$
Applying the power rule and chain rule:
$$f\text{'}\left(x\right)=12{\sin }^{2}x\cos x-12\sin x\cos x+12\cos x$$

We can factor out $12\cos x$ from the expression:
$$f\text{'}\left(x\right)=12\cos x({\sin }^{2}x-\sin x+1)$$

Now, let us analyze the sign of each factor in the derivative:
1. Consider the quadratic term in the parentheses: $q\left(x\right)={\sin }^{2}x-\sin x+1$.
Letting $u=\sin x$, where $u\in [-1,1]$, the quadratic expression becomes $u^2-u+1$.
The discriminant of this quadratic is $D={(-1)}^2-4\left(1\right)\left(1\right)=1-4=-3$.
Since the discriminant is negative and the leading coefficient is positive (1 > 0), the expression ${\sin }^{2}x-\sin x+1$ is strictly positive for all real values of $x$.

Therefore, the sign of $f\text{'}\left(x\right)$ is determined entirely by the term $\cos x$.
Specifically:
• If $\cos x>0$, then $f\text{'}\left(x\right)>0$ (strictly increasing).
• If $\cos x<0$, then $f\text{'}\left(x\right)<0$ (strictly decreasing).

Let us evaluate the interval $[-\pi /2,\pi /2]$:
In the interval $(-\pi /2,\pi /2)$ (which represents the fourth and first quadrants), the cosine function is strictly positive, i.e., $\cos x>0$.
Thus, the function is strictly increasing in this interval, and strictly decreasing in the regions where $\cos x<0$, such as $(\pi /2,3\pi /2)$.
Following the specific provided correct option, the function behaves monotonically based on the sign of $\cos x$.