If \frac{\cos^{2}\theta-3\cos\theta+2}{\sin^{2}\theta}=1, where 0^{\circ} \lt \theta \lt 90^{\circ}, then what is \sin^{2}\theta+\cos\theta equal to?

Explanation

Substitute \sin^2\theta = 1 - \cos^2\theta, so \cos^2\theta - 3\cos\theta + 2 = 1 - \cos^2\theta. Rearranging gives 2\cos^2\theta - 3\cos\theta + 1 = 0, factoring to (2\cos\theta - 1)(\cos\theta - 1) = 0. Since \theta \neq 0, \cos\theta = \frac{1}{2} \implies \theta = 60^{\circ}. Thus, \sin^2 60^{\circ} + \cos 60^{\circ} = \frac{3}{4} + \frac{1}{2} = \frac{5}{4}.

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