Cos2x+sin2x+2cos^2x=0 Решите уравнение

1. Воспользуемся формулами:

sin2α = 2sinα * cosα; cos2α = cos^2α - sin^2α; cos2x + sin2x + 2cos^2x = 0; cos^2x - sin^2x + 2sinx * cosx + 2cos^2x = 0; 3cos^2x - sin^2x + 2sinx * cosx = 0.

2. Разделим на cos^2x:

3 - tg^2x + 2tgx = 0; tg^2x - 2tgx - 3 = 0.

3. Четный второй коэффициент:

a = 1; c = - 3; b = - 2; b/2 = - 1; D/4 = (b/2) ^2 - ac; D/4 = (-1) ^2 - 1 * (-3) = 1 + 3 = 4; tgx = (-b/2 ± √ (D/4)) / a; tgx = (1 ± √4) / 1 = 1 ± 2;

1) tgx = 1 - 2 = - 1;

x = - π/4 + πk, k ∈ Z;

2) tgx = 1 + 2 = 3;

x = arctg3 + πk, k ∈ Z.

Ответ: - π/4 + πk; arctg3 + πk, k ∈ Z.