2cos^2 (4x) - 6cos^2 (2x) + 1=0

2cos^2 (4x) - 6cos^2 (2x) + 1 = 0;

2 (2cos^2 (2x) - 1) ^2 - 6cos^2 (2x) + 1 = 0;

2 (4cos^4 (2x) - 4cos^2 (2x) + 1) - 6cos^2 (2x) + 1 = 0;

8cos^4 (2x) - 14cos^2 (2x) + 3 = 0;

Сделаем замену переменных: cos^2 (2x) = t; 0 ≤ t ≤ 1;

8t^2 - 14t + 3 = 0;

D = 196 - 96 = 100;

t1 = (14 + 10) / 16 = 3/2 > 1 - не удовлетворяет условиям;

t2 = (14 - 10) / 16 = 1/4;

cos^2 (2x) = 1/4; --> cos2x = + 1/2; cos2x = - 1/2;

cos2x = 1/2; 2x = ± п/3 + 2 пn, n∈Z; x=± п/6 + пn, n∈Z;

cos2x = - 1/2; 2x = ±2π/3 + 2πk; k∈Z; x = ± π/3 + πk; k∈Z.

Ответ: x=± п/6 + пn, n∈Z; x = ± π/3 + πk; k∈Z.