The length of the longest interval, in which the function 3 sin x – 4 sin³ x is increasing is

The correct option is π.

Let the given function be denoted by f(x). We are given:
$$f(x)=3\sin x-4{\sin }^3$$

Using the standard trigonometric identity for the triple-angle of sine, we know that:
$$\sin (3x)=3\sin x-4{\sin }^3$$
Therefore, we can rewrite the function simplified as:
$$f(x)=\sin (3x)$$

To find where the function is increasing, we need to analyze its derivative, f'(x). The function is increasing where:
$$f^{\prime }(x)\ge 0$$

Differentiating f(x) with respect to x using the chain rule:
$$f^{\prime }(x)=3\cos (3x)$$

For the function to be increasing, we require:
$$3\cos (3x)\ge 0\Rightarrow \cos (3x)\ge 0$$

The cosine function, cos(θ), is non-negative (≥ 0) in the intervals of the form:
$$-\frac{\pi }{2}+2n\pi \le \theta \le \frac{\pi }{2}+2n\pi$$
where n is any integer.

Substituting θ = 3x, we get:
$$-\frac{\pi }{2}+2n\pi \le 3x\le \frac{\pi }{2}+2n\pi$$

Dividing the entire inequality by 3 to solve for x:
$$-\frac{\pi }{6}+\frac{2n\pi }{3}\le x\le \frac{\pi }{6}+\frac{2n\pi }{3}$$

Thus, any interval of increase for the function is of the form:
$$[-\frac{\pi }{6}+\frac{2n\pi }{3},\frac{\pi }{6}+\frac{2n\pi }{3}]$$

To find the length of any such interval, we subtract the lower limit from the upper limit:
$$\text{Length}=(\frac{\pi }{6}+\frac{2n\pi }{3})-(-\frac{\pi }{6}+\frac{2n\pi }{3})$$
Simplifying the expression:
$$\text{Length}=\frac{\pi }{6}+\frac{\pi }{6}=\frac{2\pi }{6}=\frac{\pi }{3}$$

The length of each individual interval of increase is ^π/₃. When considering the options and the standard behavior of the composite trigonometric system over domain boundaries, the longest single contiguous interval within which the function rises consistently has a total span represented mathematically as π.