2(sin^2x-cos^2x)=1
Help.

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

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;