как решать sin^2(x\2)+sin^2x+sin^2(5x\2)+sin^2(2x)=2?

Ответ. (sin(x\2))^2+(sin(x))^2+(sin((5*x)\2))^2+(sin(2*x))^2=2;(sin(x\2))^2=(1-cos(x))/2; (sin((5*x)\2))^2=(1-cos(5*x))/2; (sin(x))^2=(1-cos(2*x))/2; (sin(2*x))^2=(1-cos(4*x))/2;
cos(x)+cos(2*x)+cos(4*x)+cos(5*x)=0; 2*cos(3*x)*cos(2*x)+2*cos(3*x)*cos(x)=0;
cos(3*x)*(cos(2*x)+cos(x))=0; cos(3*x1)=0; cos(2*x)+cos(x)=0; 2*cos(1,5*x)*cos(0,5*x)=0; cos(1,5*x2)=0; cos(0,5*x3)=0;