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Find all solutions of the equation. sin x = -(radical3)/2
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\[\sin x = -\frac{\sqrt 3}{2}\]
\(\bf sin(x)=-\cfrac{\sqrt{3}}{2}\implies sin^{-1}[sin(x)]=sin^{-1}\left(-\cfrac{\sqrt{3}}{2}\right)\\ \quad \\ \measuredangle x==sin^{-1}\left(-\cfrac{\sqrt{3}}{2}\right)\)
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