TrigonometryEasyTrigonometric Equations ShareTrigonometric Equations · Easy problemSolve 2cosx=32\cos x = \sqrt{3}2cosx=3 on [0,2π)[0,2\pi)[0,2π).Aπ6, 11π7\dfrac{\pi}{6},\ \dfrac{11\pi}{7}6π, 711πB6π, 11π6\dfrac{6}{\pi},\ \dfrac{11\pi}{6}π6, 611πCπ6, 11π6\dfrac{\pi}{6},\ \dfrac{11\pi}{6}6π, 611πDπ12, 11π6\dfrac{\pi}{12},\ \dfrac{11\pi}{6}12π, 611πCheck Show answer Walk through it step by step Practice more like this Review this concept