If \cot^{2}\theta-3\sqrt{3}\cot \theta+6=0, where \frac{\pi}{6}\leq\theta\lt \frac{\pi}{2}, then what is a value of \sin \theta+\cos 2\theta?

Explanation

Solving the quadratic for \cot\theta gives \cot\theta = \sqrt{3} or 2\sqrt{3}. Taking \cot\theta = \sqrt{3}, we get \theta = 30^\circ. Evaluating \sin 30^\circ + \cos 60^\circ = 0.5 + 0.5 = 1.

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