1) 4sin^2x=3sinxcosx+cos^2x2) cos2x*cos4x=-1

1)

4sin^2x = 3sinx * cosx + cos^2x; 4sin^2x - 3sinx * cosx - cos^2x = 0; 4tg^2x - 3tgx - 1 = 0; D = 3^2 + 4 * 4 = 25 = 5^2; tgx = (3 ± 5) / 8;

a) tgx = (3 - 5) / 8 = - 2/8 = - 1/4;

x = - arctg (1/4) + πk, k ∈ Z;

b) tgx = (3 + 5) / 8 = 8/8 = 1;

x = π/4 + πk, k ∈ Z.

2) Возможны два случая:

cos2x * cos4x = - 1;

a)

{cos2x = - 1;

{cos4x = 1; {2x = π + 2πk, k ∈ Z;

{4x = 2πk, k ∈ Z; {x = π/2 + πk, k ∈ Z;

{x = πk/2, k ∈ Z; π/2 + πk, k ∈ Z.

b)

{cos2x = 1;

{cos4x = - 1; {2x = 2πk, k ∈ Z;

{4x = π + 2πk, k ∈ Z; {x = πk, k ∈ Z;

{x = π/4 + πk/2, k ∈ Z - нет решений.

Ответ:

1) - arctg (1/4) + πk; π/4 + πk, k ∈ Z; 2) π/2 + πk, k ∈ Z.