А) sin2x=cos2x б) 2sin^2*2x-5sin*2x*cos*2x+2cos^2*2x=0

а) Разделим обе части уравнения на cos2x:

sin2x = cos2x; tg2x = 1; 2x = π/4 + πk, k ∈ Z; x = π/8 + πk/2, k ∈ Z.

б) Разделим уравнение на cos^2 (2x):

2sin^2 (2x) - 5sin2x * cos2x + 2cos^2 (2x) = 0; 2tg^2 (2x) - 5tg2x + 2 = 0; D = 5^2 - 4 * 2 * 2 = 25 - 16 = 9; tg2x = (5 ± √9) / 4 = (5 ± 3) / 4;

1) tg2x = (5 - 3) / 4 = 2/4 = 1/2;

2x = arctg (1/2) + πk, k ∈ Z; x = 1/2 * arctg (1/2) + πk, k ∈ Z;

2) tg2x = (5 + 3) / 4 = 8/4 = 2;

2x = arctg (2) + πk, k ∈ Z. x = 1/2 * arctg (2) + πk, k ∈ Z.

Ответ:

а) π/8 + πk/2, k ∈ Z; б) 1/2 * arctg (1/2); 1/2 * arctg (2) + πk, k ∈ Z.