The number of values of x satisfying sin 4x = | vedclass

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(B) Given $\cos 3x = \sin 4x$. We can write this as $\cos 3x = \cos \left(\frac{\pi}{2} - 4xight)$. Using the general solution $\cos 	heta = \cos \

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$1$

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(B) Given $\cos 3x = \sin 4x$. We can write this as $\cos 3x = \cos \left(\frac{\pi}{2} - 4xight)$. Using the general solution $\cos 	heta = \cos \alpha \Rightarrow 	heta = 2n\pi \pm \alpha$,where $n \in \mathbb{Z}$: Case $1$: $3x = 2n\pi + \left(\frac{\pi}{2} - 4xight) $ $\Rightarrow 7x = 2n\pi + \frac{\pi}{2} $ $\Rightarrow x = \frac{2n\pi}{7} + \frac{\pi}{14}$. For $n=0$,$x = \frac{\pi}{14} \approx 0.224$ (which is in $(-\frac{\pi}{6}, \frac{\pi}{6})$ since $\frac{\pi}{14} For $n=-1$,$x = -\frac{2\pi}{7} + \frac{\pi}{14} = -\frac{3\pi}{14} Case $2$: $3x = 2n\pi - \left(\frac{\pi}{2} - 4xight) $ $\Rightarrow 3x = 2n\pi - \frac{\pi}{2} + 4x $ $\Rightarrow -x = 2n\pi - \frac{\pi}{2} $ $\Rightarrow x = -2n\pi + \frac{\pi}{2}$. For $n=0$,$x = \frac{\pi}{2} > \frac{\pi}{6}$. For $n=1$,$x = -2\pi + \frac{\pi}{2} = -\frac{3\pi}{2} Wait,checking the interval again,if the interval is $(-\frac{\pi}{6}, \frac{\pi}{6})$,let's re-evaluate $x = \frac{2n\pi}{7} + \frac{\pi}{14}$. For $n=0$,$x = \frac{\pi}{14}$. Are there others? Let's check $x = -\frac{\pi}{14}$? No. Actually,$\sin 4x = \cos 3x \Rightarrow \sin 4x = \sin(\frac{\pi}{2} - 3x)$. $4x = n\pi + (-1)^n(\frac{\pi}{2} - 3x)$. If $n=0$,$4x = \frac{\pi}{2} - 3x$ $\Rightarrow 7x = \frac{\pi}{2}$ $\Rightarrow x = \frac{\pi}{14}$. If $n=1$,$4x = \pi - (\frac{\pi}{2} - 3x) = \frac{\pi}{2} + 3x \Rightarrow x = \frac{\pi}{2}$. If $n=-1$,$4x = -\pi - (\frac{\pi}{2} - 3x) = -\frac{3\pi}{2} + 3x \Rightarrow x = -\frac{3\pi}{2}$. Thus,only $x = \frac{\pi}{14}$ is in the interval $(-\frac{\pi}{6}, \frac{\pi}{6})$. There is $1$ value.

The number of values of $x$ satisfying $\sin 4x = \cos 3x$ in the interval $\left(-\frac{\pi}{6}, \frac{\pi}{6}\right)$ is:

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