PrecalculusHard SharePrecalculus · Hard problemSolve sin2x=sinx\sin2x=\sin xsin2x=sinx on [0,2π)[0,2\pi)[0,2π).A0, π3, π, 5π40,\ \dfrac{\pi}{3},\ \pi,\ \dfrac{5\pi}{4}0, 3π, π, 45πB−0, π3, π, 5π3-0,\ \dfrac{\pi}{3},\ \pi,\ \dfrac{5\pi}{3}−0, 3π, π, 35πC0, π3, π, 5π30,\ \dfrac{\pi}{3},\ \pi,\ \dfrac{5\pi}{3}0, 3π, π, 35πD0, 3π, π, 5π30,\ \dfrac{3}{\pi},\ \pi,\ \dfrac{5\pi}{3}0, π3, π, 35πCheck Show answer Walk through it step by step Practice more like this