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annakornilova2
@annakornilova2
July 2022
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1)cos2x/sin2x+1=0
2) (1+sinx)*(1- tgx/2)=0
3)sin4x*(sin(x+п/4)-1)=0 решите плиз
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Djash
1)tg2x+1=0
tg2x=-1
2х=arctg(-1)+πn
2х=-π/4+πn
х=-π/8+πn/2 n
∈ Z
2)1+sinx=0 и 1-tgx/2=0
sinx=-1 tgx=1/2
х=-π/2 + 2πn, nЄZ х=arctg(-1)+πn
х=-π/4+πn, nЄZ
3)sinx4x=0 и sin(x+π/4)-1=0
x=πn/4, nЄZ sin(x+π/4)=1
x=π\2+2πn, nЄZ
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Djash
Пожалуйста)
annakornilova2
сидишь вк?
Djash
Ага
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ок
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annakornilova2
добавь меня сам пожалуйста
annakornilova2
ты Рустам?)
Djash
он самый
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Answers & Comments
tg2x=-1
2х=arctg(-1)+πn
2х=-π/4+πn
х=-π/8+πn/2 n ∈ Z
2)1+sinx=0 и 1-tgx/2=0
sinx=-1 tgx=1/2
х=-π/2 + 2πn, nЄZ х=arctg(-1)+πn
х=-π/4+πn, nЄZ
3)sinx4x=0 и sin(x+π/4)-1=0
x=πn/4, nЄZ sin(x+π/4)=1
x=π\2+2πn, nЄZ