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vadimz10
@vadimz10
August 2022
2
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решите уравнение 1/(tg^2x) - 2/(tgx
) - 3 = 0 укажите корни принадлежащие отрезку (2п, 7п/2)
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01729890
1/(tg^2 x) - 2/tgx - 3 = 0
1 - 2tgx - 3tg^2 x = 0
3tg^2 x + 2tgx - 1 = 0
3tgx -1
1tgx 1
(3tgx - 1)(tgx+1)=0
tgx = 1/3
x = arctg 1/3 + pk
tgx = -1
x = arctg(-1) + pk
3 votes
Thanks 4
mariyagurtikow
Tgx=1/3;
thx=-1;
x=arctg(-1)+pk.
1 votes
Thanks 3
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Answers & Comments
1 - 2tgx - 3tg^2 x = 0
3tg^2 x + 2tgx - 1 = 0
3tgx -1
1tgx 1
(3tgx - 1)(tgx+1)=0
tgx = 1/3
x = arctg 1/3 + pk
tgx = -1
x = arctg(-1) + pk
thx=-1;
x=arctg(-1)+pk.