TrigonometryMediumTrigonometric Equations ShareTrigonometric Equations · Medium problemSolve 2sin2x−sinx−1=02\sin^2 x - \sin x - 1 = 02sin2x−sinx−1=0 on [0,2π)[0,2\pi)[0,2π).Check Show answer Walk through it step by step Practice more like this Review this concept