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# help

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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.)

Oct 3, 2019

$$4\cos^2(2\theta-\pi)=3\\ \cos(2\theta-\pi) = \pm \dfrac{\sqrt{3}}{2}\\ \text{We need the 3 smallest positive values of \theta so we have}\\ 2\theta - \pi = -\dfrac {5\pi}{6}, -\dfrac{\pi}{6}, \dfrac{\pi}{6}\\ 2\theta = \dfrac \pi 6,~\dfrac{5\pi}{6},~\dfrac{7\pi}{6}\\ \theta = \dfrac{\pi}{12},~\dfrac{5\pi}{12},~\dfrac{7\pi}{12}$$