90°
60°
45°
30°
Correct answer is B
2\sin^{2} \theta = 1 + \cos \theta
2 ( 1 - \cos^{2} \theta) = 1 + \cos \theta
2 - 2\cos^{2} \theta = 1 + \cos \theta
0 = 1 - 2 + \cos \theta + 2\cos^{2} \theta
2\cos^{2} \theta + \cos \theta - 1 = 0
Factorizing, we have
(\cos \theta + 1)(2\cos \theta - 1) = 0
Note: In the range, 0° \leq \theta \leq 90°, all trig functions are positive, so we consider
2\cos \theta = 1 \implies \cos \theta = \frac{1}{2}
\theta = 60°.
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