Sin в квадрате x+sin x*cosx-2 cos в квадрате x=0

sin^2x+sinx*cosx-2cos^2x=0 sin^2x+sinxcosx -2 +2sin^2x=0 3sin^2x +sinxcosx -2 =0 3sin^2x +sinxcosx -2sin^2x - 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 x1 = -1+3/2 =1; x2= -1-3/2 = -2; tgx = 1                   tgx=-2 tgx = 1: x=arctg1 + pn, n e z ; x= p/4 +pn, nez; tgx = -2 : x= arctg ( -2 ) + pk, k e z; ответ : p/4 + pn, nez ;  arctg (-2) +pk, k e z. Фууух :)