Вопрос пользователя:
sin^2x-2sinx*cosx-3cos^2x=0
Илюха отвечает:
делим на cos^2x
tg^2x – 2tgx – 3 = 0
tgx = a
a^2 -2a -3 = 0
2±√4+12 /2
a1 = 2+4/2 = 3
a2 = 2-4/2 = -1
1) tgx = 3
x = arctg3 + πn
2) tgx = -1
x = -π/4 +πn
sin^2x-2sinx*cosx-3cos^2x=0
делим на cos^2x
tg^2x – 2tgx – 3 = 0
tgx = a
a^2 -2a -3 = 0
2±√4+12 /2
a1 = 2+4/2 = 3
a2 = 2-4/2 = -1
1) tgx = 3
x = arctg3 + πn
2) tgx = -1
x = -π/4 +πn