The length of a longest interval in which the function 3sin x -4 sin^3(x) is increasing, is

Key Concepts Needed

1. Derivative & Monotonicity
   A function $f(x)$ is increasing where its derivative $f'(x) > 0$.

2. Trigonometric Identity
   $\sin(3x) = 3\sin x - 4\sin^3 x$
   This lets us replace the messy cubic sine expression with a neat single sine term.

3. Sign of Cosine
   The derivative of $\sin(3x)$ is $3\cos(3x)$.
   Therefore $\sin(3x)$ is increasing where $\cos(3x) > 0$ and decreasing where $\cos(3x) < 0$.

4. Intervals Where $\cos(\theta)$ is Positive
   In one full cycle ($0$ to $2\pi$), $\cos\theta$ stays positive from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. Shifting by multiples of $2\pi$ gives all such intervals.

5. Scaling the Interval
   Replacing $\theta$ with $3x$ compresses the interval by a factor of $3$.

Chain of Thought to Crack the Problem

1. Recognise the Identity
   Spot that $3\sin x - 4\sin^3 x$ equals $\sin(3x)$. This removes algebraic clutter.

2. Differentiate
   $f(x)=\sin(3x)$
   $f'(x)=3\cos(3x)$

3. Set Derivative Positive
   $3\cos(3x)>0 \;\Longrightarrow\; \cos(3x)>0$

4. Locate Positive Cosine Zones
   $3x \in \left(-\frac{\pi}{2}+2k\pi,\; \frac{\pi}{2}+2k\pi\right),\; k\in\mathbb{Z}$

5. Translate Back to $x$
   $x \in \left(-\frac{\pi}{6}+\frac{2k\pi}{3},\; \frac{\pi}{6}+\frac{2k\pi}{3}\right)$

6. Find Interval Length
   Length $= \frac{\pi}{6} - \left(-\frac{\pi}{6}\right) = \frac{\pi}{3}$.

Thus the longest continuous stretch on which the function rises is of length $\frac{\pi}{3}$.

🎈 Imagine a Roller-Coaster!

You know how a roller-coaster first climbs up a hill and then comes down?
If we draw the height of the coaster on paper, the up-hill part is where the line goes up.
In maths, when a graph goes up, we say the function is increasing.

Our roller-coaster track here is the wiggly curve $3\sin x - 4\sin^3 x$.
Surprise! That wiggly curve is actually the same as $\sin(3x)$ (it’s a famous trigonometry identity).

So the question becomes:
👉 "For how wide a stretch is $\sin(3x)$ climbing upward before it starts sliding down again?"

Sine waves go up whenever their slope (derivative) is positive.
For $\sin(3x)$ the slope is $3\cos(3x)$.
$\cos(3x)$ is positive in the middle region of each wave-hump. That middle region covers exactly one-third of the usual sine interval, giving a longest increasing stretch of length $\frac{\pi}{3}$.

So, like the longest straight climb on the roller-coaster, the answer is a distance of $\frac{\pi}{3}$ along the $x$-axis.

Step-by-Step Solution

1. Recognise the Identity $3\sin x - 4\sin^3 x = \sin(3x)$

2. Define the Function
   $f(x)=\sin(3x)$

3. Differentiate
   $f'(x)=\frac{d}{dx}[\sin(3x)] = 3\cos(3x)$

4. Set Derivative Positive for Increasing Interval
   $f'(x) > 0 \;\Longrightarrow\; 3\cos(3x) > 0 \;\Longrightarrow\; \cos(3x) > 0$

5. Solve for $x$
   $\cos\theta>0$ when $\theta \in \left(-\frac{\pi}{2}+2k\pi,\; \frac{\pi}{2}+2k\pi\right),\; k\in\mathbb{Z}$ Put $\theta = 3x$: $3x \in \left(-\frac{\pi}{2}+2k\pi,\; \frac{\pi}{2}+2k\pi\right)$ Divide by $3$: $x \in \left(-\frac{\pi}{6}+\frac{2k\pi}{3},\; \frac{\pi}{6}+\frac{2k\pi}{3}\right)$

6. Find Length of Each Increasing Interval
   Length $= \frac{\pi}{6} - \left(-\frac{\pi}{6}\right) = \frac{\pi}{3}$

7. Answer
   The longest interval over which the function is increasing has length:
   $\boxed{\dfrac{\pi}{3}}$