If \(8\sin\theta - \cos\theta = 4\), where \(\dfrac{\pi}{6} < \theta < \dfrac{\pi}{3}\), then what is \(\sin\theta\) equal to?
\(\dfrac{3}{5}\)
From \(8\sin\theta - \cos\theta = 4\), \(\cos\theta = 8\sin\theta - 4\). Substituting in \(\sin^{2}\theta+\cos^{2}\theta=1\) gives \(65\sin^{2}\theta - 64\sin\theta + 15 = 0\), so \(\sin\theta = \dfrac{64\pm\sqrt{64^{2}-4(65)(15)}}{130} = \dfrac{64\pm14}{130}\), giving \(\sin\theta = \dfrac{3}{5}\) or \(\dfrac{5}{13}\). Since \(\dfrac{\pi}{6}<\theta<\dfrac{\pi}{3}\), \(\sin\theta\) must lie between \(\sin 30^{\circ}=0.5\) and \(\sin 60^{\circ}\approx0.866\). Only \(\dfrac{3}{5}=0.6\) satisfies this, so \(\sin\theta = \dfrac{3}{5}\).
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