If sin(2x) = frac{sqrt{5}-1}{4},then x = frac | vedclass

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(C) Given the equation: $\sin(2x) = \frac{\sqrt{5}-1}{4}$.We know that $\sin(18^\circ) = \sin(\frac{\pi}{10}) = \frac{\sqrt{5}-1}{4}$.So,$\sin(2x) = \

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$\frac{\pi}{20}$

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(C) Given the equation: $\sin(2x) = \frac{\sqrt{5}-1}{4}$.We know that $\sin(18^\circ) = \sin(\frac{\pi}{10}) = \frac{\sqrt{5}-1}{4}$.So,$\sin(2x) = \sin(\frac{\pi}{10})$.The general solution for $\sin(	heta) = \sin(\alpha)$ is $	heta = n\pi + (-1)^n\alpha$.Applying this to $2x = \frac{\pi}{10}$:$2x = n\pi + (-1)^n(\frac{\pi}{10})$.Dividing by $2$:$x = \frac{n\pi}{2} + (-1)^n(\frac{\pi}{20})$.Comparing this with the given form $x = \frac{n\pi}{2} + (-1)^n(m)$,we find $m = \frac{\pi}{20}$.Thus,option $C$ is correct.

If $\sin(2x) = \frac{\sqrt{5}-1}{4}$,then $x = \frac{n\pi}{2} + (-1)^n(m)$,where $n \in \mathbb{Z}$. Find $m$.

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