What is the length of the longest interval in | vedclass

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(A) Given the function $f(x) = 3\sin x - 4\sin^3 x$. Using the trigonometric identity,we know that $3\sin x - 4\sin^3 x = \sin(3x)$. So,$f(x) = \sin(3

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$\pi /3$

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(A) Given the function $f(x) = 3\sin x - 4\sin^3 x$. Using the trigonometric identity,we know that $3\sin x - 4\sin^3 x = \sin(3x)$. So,$f(x) = \sin(3x)$. For the function to be increasing,its derivative must be positive: $f'(x) > 0$. $f'(x) = \frac{d}{dx}(\sin(3x)) = 3\cos(3x)$. We require $3\cos(3x) > 0$,which means $\cos(3x) > 0$. The cosine function is positive in the interval $(-\pi/2, \pi/2)$. Therefore,$-\frac{\pi}{2} Dividing by $3$,we get $-\frac{\pi}{6} The length of this interval is $\frac{\pi}{6} - (-\frac{\pi}{6}) = \frac{2\pi}{6} = \frac{\pi}{3}$.

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

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