The length of the longest interval in which t | vedclass

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(A) Given $f(x) = 3 \sin x - 4 \sin^3 x$. Using the identity $\sin 3x = 3 \sin x - 4 \sin^3 x$,we have $f(x) = \sin 3x$. To find the interval where $f

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$\frac{\pi}{3}$

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(A) Given $f(x) = 3 \sin x - 4 \sin^3 x$. Using the identity $\sin 3x = 3 \sin x - 4 \sin^3 x$,we have $f(x) = \sin 3x$. To find the interval where $f(x)$ is increasing,we find the derivative $f'(x) = 3 \cos 3x$. For $f(x)$ to be increasing,$f'(x) \geq 0$,which implies $3 \cos 3x \geq 0$,or $\cos 3x \geq 0$. The cosine function is non-negative in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Thus,$-\frac{\pi}{2} \leq 3x \leq \frac{\pi}{2}$. Dividing by $3$,we get $-\frac{\pi}{6} \leq x \leq \frac{\pi}{6}$. The length of this interval is $\frac{\pi}{6} - (-\frac{\pi}{6}) = \frac{2\pi}{6} = \frac{\pi}{3}$.

The length of the longest interval in which the function $f(x) = 3 \sin x - 4 \sin^3 x$ is increasing is:

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