The length of the interval in which the funct | vedclass

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(A) Given $f(x) = 3 \sin x - 4 \sin^3 x$. Using the trigonometric identity $\sin 3x = 3 \sin x - 4 \sin^3 x$,we have $f(x) = \sin 3x$. For $f(x)$ to b

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$\pi /3$

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(A) Given $f(x) = 3 \sin x - 4 \sin^3 x$. Using the trigonometric identity $\sin 3x = 3 \sin x - 4 \sin^3 x$,we have $f(x) = \sin 3x$. For $f(x)$ to be an increasing function,its derivative $f'(x)$ must be greater than $0$. $f'(x) = \frac{d}{dx}(\sin 3x) = 3 \cos 3x$. For $f(x)$ to be increasing,$3 \cos 3x > 0$,which implies $\cos 3x > 0$. This occurs when $-\frac{\pi}{2} Dividing by $3$,we get $-\frac{\pi}{6} The length of this interval is $\frac{\pi}{6} - (-\frac{\pi}{6}) = \frac{2\pi}{6} = \frac{\pi}{3}$.

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

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