TrigonometryEasyTrigonometric Equations ShareTrigonometric Equations · Easy problemSolve sinx=12\sin x = \dfrac{1}{2}sinx=21 on [0,2π)[0,2\pi)[0,2π).Aπ6, 5π7\dfrac{\pi}{6},\ \dfrac{5\pi}{7}6π, 75πBπ6, 5π6\dfrac{\pi}{6},\ \dfrac{5\pi}{6}6π, 65πCπ12, 5π6\dfrac{\pi}{12},\ \dfrac{5\pi}{6}12π, 65πD6π, 5π6\dfrac{6}{\pi},\ \dfrac{5\pi}{6}π6, 65πCheck Show answer Walk through it step by step Practice more like this Review this concept