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 = -2
x = 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.