Given that \(\sin x = \frac{-\sqrt{3}}{2}\) and \(\cos x > 0\), find x.
In order to do this, simply find the option in the range where only the cos is +ve. This occurs in the range \(270° \leq x \leq 360°\).
Check: \(\sin 300 = - \sin 60 = \frac{-\sqrt{3}}{2}\)
\(\cos 300 = \cos 60 = \frac{1}{2}\)
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