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Answer: (B) \(\dfrac{\pi}{2}\)

Write \(\sin^2 2\theta = 1 - \cos^2 2\theta\) and let \(c = \cos^2 2\theta\):

\[1 - c + c^2 = \frac{3}{4}\]

\[c^2 - c + \frac{1}{4} = 0 \;\Rightarrow\; \left(c - \tfrac{1}{2}\right)^2 = 0\]

So \(\cos^2 2\theta = \frac{1}{2}\), which means \(\cos 4\theta = 2\cos^2 2\theta - 1 = 0\).

For \(\theta \in (0, \frac{\pi}{2})\), \(4\theta \in (0, 2\pi)\), so \(4\theta = \frac{\pi}{2}\) or \(\frac{3\pi}{2}\).

That gives \(\theta = \frac{\pi}{8}\) and \(\frac{3\pi}{8}\).

Sum \(= \dfrac{\pi}{8} + \dfrac{3\pi}{8} = \dfrac{\pi}{2}\).

Doubtly AI · 23 Oct 2026 AI-verified · Oct 2026 Report
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