1). 2cosx-√2=0; 2). sin2x+2sin^2x=2 cox^2x; 3). sin^2x+2cosx=-2

1)

2cosx - √2 = 0;

2cosx = √2;

cosx = √2/2;

x = ±π/4 + 2πk, k ∈ Z.

2)

sin2x + 2sin²x = 2cos²x;

sin2x = 2cos²x - 2sin²x;

sin2x = 2 (cos²x - sin²x);

sin2x = 2cos2x;

sin2x/cos2x = 2;

tgx = 2;

x = arctg2 + πk, k ∈ Z.

3)

sin²x + 2cosx = - 2;

1 - cos²x + 2cosx + 2 = 0;

-cos²x + 2cosx + 3 = 0;

cos²x - 2cosx - 3 = 0;

D/4 = 1² + 3 = 4;

cosx = 1 ± √4 = 1 ± 2;

1) cosx = 1 - 2 = - 1;

x = π + 2πk, k ∈ Z;

2) cosx = 1 + 2 = 3 > 1 - не имеет решения.

Ответ:

1) ±π/4 + 2πk, k ∈ Z.

2) arctg2 + πk, k ∈ Z.

3) π + 2πk, k ∈ Z.