TrigonometryEasyTrigonometric Equations ShareTrigonometric Equations · Easy problemSolve 2sin2x+cosx−1=02\sin^2 x + \cos x - 1 = 02sin2x+cosx−1=0 on [0,2π)[0,2\pi)[0,2π). (Hint: sin2x=1−cos2x\sin^2 x = 1-\cos^2 xsin2x=1−cos2x.)A0, 2π3, 4π30,\ \dfrac{2\pi}{3},\ \dfrac{4\pi}{3}0, 32π, 34πB0, 2π3, 4π40,\ \dfrac{2\pi}{3},\ \dfrac{4\pi}{4}0, 32π, 44πC0, 32π, 4π30,\ \dfrac{3}{2\pi},\ \dfrac{4\pi}{3}0, 2π3, 34πD−0, 2π3, 4π3-0,\ \dfrac{2\pi}{3},\ \dfrac{4\pi}{3}−0, 32π, 34πCheck Show answer Walk through it step by step Practice more like this Review this concept