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andromeda31
@andromeda31
August 2022
2
15
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a) Решите уравнение 3 cos^2x+cosx-4=0
б)Укажите корни, принадлежащие отрезку [п/2;3п]
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skyne8
Verified answer
=3(2(cosx)^2 -1)+cosx-4=0,
6cos^2(x)+cosx-7=0,
(cosx -1)(cosx -14/6)=0, cosx=1,
x=2п
3 votes
Thanks 2
jazka001
Отметим cosх-у и будем решить уравнения 3у^2+у-4=0 D=1+48=49 y1=1 y2=-4/3 теперь cosx=1 x=2pin,n €Z cosx=-4/3не сответствует ответ х=2pin
3 votes
Thanks 0
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Answers & Comments
Verified answer
=3(2(cosx)^2 -1)+cosx-4=0,6cos^2(x)+cosx-7=0,
(cosx -1)(cosx -14/6)=0, cosx=1,
x=2п