TrigonometryEasyTrigonometric Equations ShareTrigonometric Equations · Easy problemSolve sin (x2)=12\sin\!\left(\dfrac{x}{2}\right) = \dfrac{1}{2}sin(2x)=21 on [0,2π)[0,2\pi)[0,2π).Aπ3, 5π3\dfrac{\pi}{3},\ \dfrac{5\pi}{3}3π, 35πB3π, 5π3\dfrac{3}{\pi},\ \dfrac{5\pi}{3}π3, 35πCπ6, 5π3\dfrac{\pi}{6},\ \dfrac{5\pi}{3}6π, 35πDπ3, 5π4\dfrac{\pi}{3},\ \dfrac{5\pi}{4}3π, 45πCheck Show answer Walk through it step by step Practice more like this Review this concept