Cos(2x+ (π/4)) = cos(x), ⇔ cos(2x + (π/4)) - cos(x) = 0,
используем триг-ую формулу "разность косинусов":
cos(a) - cos(b) = -2*sin((a+b)/2)*sin((a-b)/2),
тогда
cos(2x + (π/4)) - cos(x) = -2*sin( (2x+(π/4)+x)/2 )*sin( (2x+(π/4) - x)/2) = 0;
-2*sin( (3x+(π/4))/2 )*sin( (x+(π/4))/2) = 0;
sin( (3x/2) + (π/8) )*sin( (x/2)+(π/8) ) = 0;
1) sin( (3x/2) + (π/8)) = 0;
или
2) sin( (x/2)+(π/8) ) = 0;
1) sin((3x/2)+(π/8)) = 0, ⇔ (3x/2)+(π/8) = πn, n∈Z,
3x/2 = -(π/8) + πn,
3x = -(π/4) + 2πn,
x = -(π/12) + (2πn/3).
2) sin( (x/2)+(π/8) ) = 0, ⇔ (x/2)+(π/8) = πm, m∈Z,
x/2 = -(π/8) + πm,
x = -(π/4) + 2πm.
Ответ. x = -(π/12) + (2πn/3), n∈Z, или x = -(π/4) + 2πm, m∈Z.