Решите уравнение 3sin^2x + sinxcosx - 2cos^2x = 0

3 * sin^2 x + sin x * cos x - 2 * cos^2 x = 0;

Разделим уравнение на cos^2 x.

3 * sin^2 x/cos^2 x + sin x * cos x/cos^2 x - 2 * cos^2 x/cos^2 x = 0;

3 * tg^2 x + tg x - 2 = 0;

D = 1^2 - 4 * 3 * (-2) = 1 + 12 * 2 = 1 + 24 = 25 = 5^2;

tg x = (-1 + 5) / (2 * 3) = 4/6 = 2/3;

tg x = (-1 - 5) / (2 * 3) = - 6/6 = - 1;

1) tg x = 2/3;

x = arctg (2/3) + pi * n, n ∈ Z;

2) tg x = - 1;

x = - pi/4 + pi * n, n ∈ Z;

Ответ: x = arctg (2/3) + pi * n и x = - pi/4 + pi * n, n ∈ Z.