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

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sin^2x+sinx*cosx-2cos^2x=0sin^2x+sinxcosx -2 +2sin^2x=03sin^2x +sinxcosx -2 =03sin^2x +sinxcosx -2sin^2x - 2cos^2x =0sin^2x +sinxcosx -2cos^2x =0  |  /cos^2xtg^2x +tgx -2 =0делаем замену:tgx = tt^2 + t -2 = 0d= 1- 4* 1 * (-2) = 9x1 = -1+3/2 =1;x2= -1-3/2 = -2;tgx = 1                   tgx=-2tgx = 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 :)