If the general solution of sin 5x = cos 2x is | vedclass

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(B) Given the equation: $\sin 5x = \cos 2x$. We can rewrite $\cos 2x$ as $\sin(\frac{\pi}{2} - 2x)$. So,$\sin 5x = \sin(\frac{\pi}{2} - 2x)$. The gene

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$\frac{2n+(-1)^n}{5+2(-1)^n}$

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(B) Given the equation: $\sin 5x = \cos 2x$. We can rewrite $\cos 2x$ as $\sin(\frac{\pi}{2} - 2x)$. So,$\sin 5x = \sin(\frac{\pi}{2} - 2x)$. The general solution for $\sin 	heta = \sin \alpha$ is $	heta = n\pi + (-1)^n \alpha$. Applying this,$5x = n\pi + (-1)^n(\frac{\pi}{2} - 2x)$. $5x = n\pi + (-1)^n \frac{\pi}{2} - (-1)^n 2x$. $5x + (-1)^n 2x = n\pi + (-1)^n \frac{\pi}{2}$. $x(5 + 2(-1)^n) = \frac{\pi}{2}(2n + (-1)^n)$. $x = \frac{\pi}{2} \cdot \frac{2n + (-1)^n}{5 + 2(-1)^n}$. Comparing this with $x = a_n \cdot \frac{\pi}{2}$,we get $a_n = \frac{2n + (-1)^n}{5 + 2(-1)^n}$.

If the general solution of $\sin 5x = \cos 2x$ is of the form $x = a_n \cdot \frac{\pi}{2}$ for $n = 0, \pm 1, \pm 2, \dots$,then $a_n =$

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