Find the sum of the three smallest positive values of \(\theta\) such that \(4\cos^2(2\theta-\pi) =3.\) (Give your answer in radians.)
+6251 \(4\cos^2(2\theta-\pi)=3\\ \cos(2\theta-\pi) = \pm \dfrac{\sqrt{3}}{2}\\ \text{We need the 3 smallest positive values of $\theta$ so we have}\\ 2\theta - \pi = -\dfrac {5\pi}{6}, -\dfrac{\pi}{6}, \dfrac{\pi}{6}\\ 2\theta = \dfrac \pi 6,~\dfrac{5\pi}{6},~\dfrac{7\pi}{6}\\ \theta = \dfrac{\pi}{12},~\dfrac{5\pi}{12},~\dfrac{7\pi}{12}\)
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