Respuesta :

Answer:

Step-by-step explanation:

[tex]2 sin^3x-sin x=0\\sinx(2sin^2x-1)=0\\either~sin x=0,x=n\pi ,where ~n~is~an~integer.\\or~2sin ^2x-1=0\\or~1-2sin^2x=0\\cos 2x=0\\2x=(2n+1)\frac{\pi }{2} \\or~x=(2n+1)\frac{\pi }{4} ,where~n~is~an~integer.[/tex]