TrigonometryMediumTrigonometric Equations ShareTrigonometric Equations · Medium problemSolve sin2x=sinx\sin 2x = \sin xsin2x=sinx on [0,2π)[0,2\pi)[0,2π). (Hint: sin2x=2sinxcosx\sin 2x = 2\sin x\cos xsin2x=2sinxcosx.)Ax=0, π3, π, 5π4x=0,\ \dfrac{\pi}{3},\ \pi,\ \dfrac{5\pi}{4}x=0, 3π, π, 45πBx=0, π3, π, 5π3x=0,\ \dfrac{\pi}{3},\ \pi,\ \dfrac{5\pi}{3}x=0, 3π, π, 35πCx=0, π3, π, 5π2x=0,\ \dfrac{\pi}{3},\ \pi,\ \dfrac{5\pi}{2}x=0, 3π, π, 25πDx=0, 3π, π, 5π3x=0,\ \dfrac{3}{\pi},\ \pi,\ \dfrac{5\pi}{3}x=0, π3, π, 35πCheck Show answer Walk through it step by step Practice more like this Review this concept