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reshator2000
@reshator2000
July 2022
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решите тригонометрическое уравнение
[tex] \frac{1}{tg^2 x} + \frac{9}{tg^2x} +8=0[/tex]
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Lesben
1/tg²x + 9/tg²x +8=0, x≠k.π/2, k∈Z
10/tg²x +8=0
tg²x bolše čem 0
10/tg²x tože bolše čem 0
10/tg²x +8 bolše čem 8
Uravnenie ne imeet rešenie
x∈∅
1/tg²x+9/tgx +8=0, x≠k.π/2,k∈Z
tgx=z
1/z²+9/z+8=0
1+9z+8z²=0,D=81-32=49,√D=√49=7
a)z1=(-9+7)/16=-2/16=-1/8
tgx=-1/8, x=arctg(-1/8) ili možno vyčislit c pomošču tablici ili kalkulatora.
b)z2=(-9-7)/16=-16/16=-1
tgx=-1, x=-π/4 +kπ, k∈Z
1 votes
Thanks 1
reshator2000
простите я ошибся
reshator2000
во второй дроби просто tg
reshator2000
спасибо огромное!!!
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Answers & Comments
10/tg²x +8=0
tg²x bolše čem 0
10/tg²x tože bolše čem 0
10/tg²x +8 bolše čem 8
Uravnenie ne imeet rešenie
x∈∅
1/tg²x+9/tgx +8=0, x≠k.π/2,k∈Z
tgx=z
1/z²+9/z+8=0
1+9z+8z²=0,D=81-32=49,√D=√49=7
a)z1=(-9+7)/16=-2/16=-1/8
tgx=-1/8, x=arctg(-1/8) ili možno vyčislit c pomošču tablici ili kalkulatora.
b)z2=(-9-7)/16=-16/16=-1
tgx=-1, x=-π/4 +kπ, k∈Z