Question

sin^4x-cos^4x=sin2x.Помогите решить!!!!!!!!!!

Answer 1

sin⁴x-cos⁴x=sin2x;
(sin²x-cos²x)(sin²x+cos²x)=sin2x;
-cos2x=sin2x;⇒cos2x≠0;
tg2x=-1;2x=-π/4+nπ;n∈Z;
x=-π/8+nπ/2;n∈Z.