Since 3sin x-4sin^3x=sin (3x) (*****) 3sin x-4sin^3x=-1 ==>sin (3x)=-1 ==>3x=3π/2 +2kπ ==>x=π/2+2kπ/3, k being an integer
(*****) sin (3x)=sin(x+2x)=sin x cos2x+cos x sin 2x =sin x(cos²x-sin²x)+2cos²x sin x =sin x(cos²x+2cos²x-sin²x) =sinx(3(1-sin²x)-sin²x) =3sin x-4sin^3x
