a) sin2x+корень (3) умножить cos2x=2sin4x. b) sin5x+cos5x=корень (2) умножить на cosx

a)

sin2x + √3cos2x = 2sin4x; 1/2 * sin2x + √3/2 * cos2x = sin4x; cos (π/3) * sin2x + sin (π/3) * cos2x = sin4x; sin (2x + π/3) - sin4x = 0; 2sin ((2x + π/3 - 4x) / 2) cos ((2x + π/3 + 4x) / 2) = 0; sin ((-2x + π/3) / 2) cos ((6x + π/3) / 2) = 0; sin (-x + π/6) cos (3x + π/6) = 0; sin (x - π/6) cos (3x + π/6) = 0; [sin (x - π/6) = 0;

[cos (3x + π/6) = 0; [x - π/6 = πk, k ∈ Z;

[3x + π/6 = π/2 + πk, k ∈ Z; [x = π/6 + πk, k ∈ Z;

[3x = π/3 + πk, k ∈ Z; [x = π/6 + πk, k ∈ Z;

[x = π/9 + πk/3, k ∈ Z.

b)

sin5x + cos5x = √2cosx; √2/2 * sin5x + √2/2 * cos5x = cosx; sin (π/4) * sin5x + cos (π/4) * cos5x = cosx; cos (5x - π/4) - cosx = 0; - 2sin ((5x - π/4 + x) / 2) sin ((5x - π/4 - x) / 2) = 0; sin ((6x - π/4) / 2) sin ((4x - π/4) / 2) = 0; sin (3x - π/8) sin (2x - π/8) = 0; [3x - π/8 = πk, k ∈ Z;

[2x - π/8 = πk, k ∈ Z; [x = π/24 + πk/3, k ∈ Z;

[x = π/16 + πk/2, k ∈ Z.

Ответ:

a) π/6 + πk; π/9 + πk/3, k ∈ Z; b) π/24 + πk/3; π/16 + πk/2, k ∈ Z.