TrigonometryEasyTrigonometric Equations ShareTrigonometric Equations · Easy problemSolve 2cos2x−cosx−1=02\cos^2 x - \cos x - 1 = 02cos2x−cosx−1=0 on [0,2π)[0,2\pi)[0,2π).A0, 2π3, 4π30,\ \dfrac{2\pi}{3},\ \dfrac{4\pi}{3}0, 32π, 34πB0, 32π, 4π30,\ \dfrac{3}{2\pi},\ \dfrac{4\pi}{3}0, 2π3, 34πC−0, 2π3, 4π3-0,\ \dfrac{2\pi}{3},\ \dfrac{4\pi}{3}−0, 32π, 34πD0, 2π3, 4π40,\ \dfrac{2\pi}{3},\ \dfrac{4\pi}{4}0, 32π, 44πCheck Show answer Walk through it step by step Practice more like this Review this concept