Ответ:π/8 + πk/4; -π/18 + 2πk/3; -5π/18 + 2πk/3, k ∈ Z.

Объяснение:cos2a = cos^2(a) - sin^2(a);

sina - sinb = 2sin((a - b)/2) * cos((a + b)/2);

sin7x + cos^2(2x) = sin^2(2x) + sinx;

sin7x - sinx + cos^2(2x) - sin^2(2x) = 0;

2sin((7x - x)/2) * cos((7x + x)/2) + cos4x = 0;

2sin3x * cos4x + cos4x = 0;

cos4x(2sin3x + 1) = 0;

[cos4x = 0;

[2sin3x + 1 = 0;

[cos4x = 0;

[sin3x = -1/2;

[4x = π/2 + πk, k ∈ Z;

[3x = -π/6 + 2πk; -5π/6 + 2πk, k ∈ Z;

[x = π/8 + πk/4, k ∈ Z;

[x = -π/18 + 2πk/3; -5π/18 + 2πk/3, k ∈ Z.

Ответ:

1)sin7x-sinx+2cos^2 2x=1

2)2( sin x + cos x) + 2 sin2x+1=0

3)корень из 3 sin x+ cos x - 2=0

4)Sin а + ctg^2 а/2=2