решить уравнение: 1) 7sin^2x-4sin2x+cos^2x=02) sin^2x-3sinx*cosx+2cos^2x=0

1)

7sin^2x - 4sin2x + cos^2x = 0; 7sin^2x - 8sinx * cosx + cos^2x = 0; 7tg^2x - 8tgx + 1 = 0; D/4 = 4^2 - 7 = 16 - 7 = 9; tgx = (4 ± √9) / 7 = (4 ± 3) / 7;

a) tgx = (4 - 3) / 7 = 1/7;

x = arctg (1/7) + πk, k ∈ Z;

b) tgx = (4 + 3) / 7 = 7/7 = 1;

x = π/4 + πk, k ∈ Z.

2)

sin^2x - 3sinx * cosx + 2cos^2x = 0; tg^2x - 3tgx + 2 = 0; D = 3^2 - 4 * 2 = 9 - 8 = 1; tgx = (3 ± 1) / 2;

a) tgx = (3 - 1) / 2 = 2/2 = 1;

x = π/4 + πk, k ∈ Z;

b) tgx = (3 + 1) / 2 = 4/2 = 2;

x = arctg (2) + πk, k ∈ Z.

Ответ:

1) arctg (1/7) + πk; π/4 + πk, k ∈ Z; 2) arctg (2) + πk; π/4 + πk, k ∈ Z.