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The length of the longest interval, in which the function $3 \sin x-4 \sin ^3 x$ is increasing, is
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$\frac{\pi}{3}$
$3 \sin x-4 \sin ^3 x=\sin 3 x$
It is increasing, when $-\pi / 2 \leq 3 x \leq \pi / 2$ i.e., $-\pi / 6 \leq x \leq \pi / 6$
The length of interval $=\left|\frac{\pi}{6}-\left(-\frac{\pi}{6}\right)\right|=\frac{\pi}{3}$
It is increasing, when $-\pi / 2 \leq 3 x \leq \pi / 2$ i.e., $-\pi / 6 \leq x \leq \pi / 6$
The length of interval $=\left|\frac{\pi}{6}-\left(-\frac{\pi}{6}\right)\right|=\frac{\pi}{3}$
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