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 :)