2(sin^2x-cos^2x)=1 Help.
Sin²x - cos²x = 1/2; sin²x = 1 - cos²x; 1 - cos²x - cos²x =1/2; 2cos²x = 1/2; cos²x = 1/4; cosx = 1/2; x = + - π/3+2πn, n∈Z;
2(1-cos^2x-cos^2x)=1 2-4cos^2x=1 -4cos^2x=-1 cos^2x=1/4 1) cosx=1/2 x=+-pi/3+2pi*n 2) cosx=-1/2 x=+-2pi/3+2pi*n Ответ: x1=pi/3+2pi*n; x2=-pi/3+2pi*n; x3=2pi/3+2pi*n; x4=-2pi/3+2pi*n