TrigonometryEasyTrigonometric Equations ShareTrigonometric Equations · Easy problemSolve sin(2x)=0\sin(2x) = 0sin(2x)=0 on [0,2π)[0,2\pi)[0,2π).Ax=0, π2, π, 3π3x=0,\ \dfrac{\pi}{2},\ \pi,\ \dfrac{3\pi}{3}x=0, 2π, π, 33πBx=0, π2, π, 3π1x=0,\ \dfrac{\pi}{2},\ \pi,\ \dfrac{3\pi}{1}x=0, 2π, π, 13πCx=0, 2π, π, 3π2x=0,\ \dfrac{2}{\pi},\ \pi,\ \dfrac{3\pi}{2}x=0, π2, π, 23πDx=0, π2, π, 3π2x=0,\ \dfrac{\pi}{2},\ \pi,\ \dfrac{3\pi}{2}x=0, 2π, π, 23πCheck Show answer Walk through it step by step Practice more like this Review this concept