PrecalculusEasy SharePrecalculus · Easy problemSolve 2sinx−1=02\sin x-1=02sinx−1=0 on [0,2π)[0,2\pi)[0,2π).A6π, 5π6\dfrac{6}{\pi},\ \dfrac{5\pi}{6}π6, 65πBπ6, 5π6\dfrac{\pi}{6},\ \dfrac{5\pi}{6}6π, 65πCπ12, 5π6\dfrac{\pi}{12},\ \dfrac{5\pi}{6}12π, 65πDπ6, 5π7\dfrac{\pi}{6},\ \dfrac{5\pi}{7}6π, 75πCheck Show answer Walk through it step by step Practice more like this