Помогите с решением. 1) sin^2x-sinx*cosx-cos^2x=3 2) 4sin^4*x+sin^2*2x=1

Ответ. 1.(sin(x))^2-sin(x)*cos(x)-(cos(x))^2=3; (sin(x))^2-sin(x)*cos(x)-(cos(x))^2=3*(sin(x))^2+3*(cos(x)0^2;; (tg(x))^2-tg(x)-1=3*(tg(x))^2+3; 2*(tg(x))^2+tg(x)+4=0; (Не имеет решений!) .
2. 4*(sin(x))^4+(sin(2*x))^2=1;4*(sin(x))^4+(sin(2*x))^2-1=0;4*(sin(x))^4-(cos(2*x))^2=0;
4*(sin(x))^4-(1-2*(sin(x))^2)^2=0; 4*(sin(x))^4-1+4*(sin(x))^2-(sin(x))^4=0; 3*(sin(x))^4-1+4*(sin(x))^2=0;
sin(x1)=0,464; sin(x2)=-0,464;