Find the general solution of sin x = -frac{sq | vedclass

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Question

(A) Given $\sin x = -\frac{\sqrt{3}}{2}$.Since $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$,we have $\sin x = -\sin \frac{\pi}{3}$.Using the identity $\s

Vedclass · Accepted Answer

$x = n\pi + (-1)^n \frac{4\pi}{3}, n \in \mathbb{Z}$

Vedclass · Answer

(A) Given $\sin x = -\frac{\sqrt{3}}{2}$.Since $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$,we have $\sin x = -\sin \frac{\pi}{3}$.Using the identity $\sin(\pi + 	heta) = -\sin 	heta$,we get $\sin x = \sin(\pi + \frac{\pi}{3}) = \sin \frac{4\pi}{3}$.The general solution for $\sin x = \sin \alpha$ is $x = n\pi + (-1)^n \alpha$,where $n \in \mathbb{Z}$.Substituting $\alpha = \frac{4\pi}{3}$,we get $x = n\pi + (-1)^n \frac{4\pi}{3}$,where $n \in \mathbb{Z}$.

Find the general solution of $\sin x = -\frac{\sqrt{3}}{2}$.

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