If 2\cos^{2}\theta+\sin\theta-2=0, 0 \lt \theta \leq \frac{\pi}{2}, then what is the value of \theta?
- A. \frac{\pi}{6} ✓
- B. \frac{\pi}{4}
- C. \frac{\pi}{3}
- D. \frac{\pi}{2}
Correct Answer: A. \frac{\pi}{6}
Explanation
Substitute \cos^2\theta = 1-\sin^2\theta into the equation: 2(1-\sin^2\theta) + \sin\theta - 2 = 0 \implies 2\sin^2\theta - \sin\theta = 0. Since \theta \gt 0, \sin\theta cannot be 0. Thus, 2\sin\theta = 1 \implies \sin\theta = \frac{1}{2}, which gives \theta = \frac{\pi}{6}.
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