Решите уравнение:

sin^2x+sinx*cosx-2cos^2x=0

Sin^2x + sinxcosx - 2cos^2x = 0 | : cos^2x

tg^2x + tgx - 2 = 0

делаем замену tgx = t;

t^2 + t - 2 = 0; d = 1 - 4 * 1 * (-2) = 9

t1 = - 1+3 / 2 = 1; t2 = - 1-3/2 = - 2;

tgx = 1; tgx = - 2x = p/4 + pk, k e z; x = arctg (-2) + pn, n e z

ответ: p/4 + pk, k e z; x = arctg (-2) + pn, n e z.