If \(2\sin^{2} \theta = 1 + \cos \theta, \leq \theta \leq find the value of \(\theta\).
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Correct Answer: Option B
Explanation:
\(2\sin^{2} \theta = 1 + \cos \theta\)
\(2 ( 1 - \cos^{2} \theta) = 1 + \cos \theta\)
\(2 - 2\cos^{2} \theta = 1 + \cos \theta\)
\(0 = 1 - 2 + \cos \theta + 2\cos^{2} \theta\)
\(2\cos^{2} \theta + \cos \theta - 1 = 0\)
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
\(2\cos \theta = 1 \implies \cos \theta = \frac{1}{2}\)
\(\theta = 60°\).
\(2\sin^{2} \theta = 1 + \cos \theta\)
\(2 ( 1 - \cos^{2} \theta) = 1 + \cos \theta\)
\(2 - 2\cos^{2} \theta = 1 + \cos \theta\)
\(0 = 1 - 2 + \cos \theta + 2\cos^{2} \theta\)
\(2\cos^{2} \theta + \cos \theta - 1 = 0\)
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
\(2\cos \theta = 1 \implies \cos \theta = \frac{1}{2}\)
\(\theta = 60°\).