Решите уравнение: 2sin^3x-cos2x=sinx

[latex]2\sin^3x-\cos2x=\sin x,\\2\sin^3x-\sin x-\cos2x=0,\\-\sin x(2\sin^2x-1)-\cos2x=0,\\-\sin x\cos2x-\cos2x=0,\\-\cos2x(\sin x+1)=0\ |\bullet(-1),\\\cos2x(\sin x+1)=0,\\\\1)\ \cos2x=0,\\2x=\frac{\pi}{2}+2\pi n,\ n\in Z,\\x=\frac{\pi}{4}+\pi n,\ n\in Z;\\\\2)\ \sin2x+1=0,\\\sin2x=-1,\\2x=-\frac{\pi}{2}+2\pi n,\ n\in Z,\\x=-\frac{\pi}{4}+\pi n,\ n\in Z.[/latex] Ответ: [latex]x=\frac{\pi}{4}+\pi n,\ x=-\frac{\pi}{4}+\pi n,\ n\in Z.[/latex] или, объединив формулы: [latex]x=\frac{\pi}{4}+\frac{\pi}{2}n,\ \ n\in Z.[/latex]