$\text{sinx}-2\text{sin}^3x=0$
$↔ \text{sinx}(1-2\text{sin}^2x)=0$
$↔ \left[ \begin{array}{l}\text{sinx}=0\\\text{sin}^2x=\dfrac{1}{2}\end{array} \right.$
$↔ \left[ \begin{array}{l}\text{sinx}=0\\\text{sinx}=±\dfrac{\sqrt[]{2}}{2}\end{array} \right.$
$↔ \left[ \begin{array}{l}x=k\pi\\x=\dfrac{\pi}{4}+k2\pi\\x=\dfrac{3\pi}{4}+k2\pi\\x=-\dfrac{\pi}{4}+k2\pi\\x=\dfrac{5\pi}{4}+k2\pi\end{array} \right.$
$↔ \left[ \begin{array}{l}x=k\pi\\x=\dfrac{\pi}{4}+k\pi\end{array} \right.$ ($k∈\mathbb{Z}$)
