TrigonometryEasyTrigonometric Equations ShareTrigonometric Equations · Easy problemSolve sinxcosx=0\sin x\cos x = 0sinxcosx=0 on [0,2π)[0,2\pi)[0,2π).Ax=0, 2π, π, 3π2x=0,\ \dfrac{2}{\pi},\ \pi,\ \dfrac{3\pi}{2}x=0, π2, π, 23πBx=0, π2, π, 3π1x=0,\ \dfrac{\pi}{2},\ \pi,\ \dfrac{3\pi}{1}x=0, 2π, π, 13πCx=0, π2, π, 3π3x=0,\ \dfrac{\pi}{2},\ \pi,\ \dfrac{3\pi}{3}x=0, 2π, π, 33πDx=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