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Austin322
@Austin322
July 2022
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Решите уравнение :
а)2cosx-корень из 2=0
б)tg2x+1=0
в)sin(x/3+пи/4)=-1
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karihmarina
А)
2cosx+1=0
2cosx=-1
cosx=-
1/2
x=
+-arcos a+2kπ, k∈Z
x=+- 2π/3+2kπ
б) tg2x+1=0
tg2x=-1
2x=-π/4+2kπ
x=-π/8+kπ
в)sin(x/3+π/4)=-1
x/3+π/4=-π/2+2kπ
x/3=-π/2+π/4+2kπ
x/3=-π/4+2kπ
x=-3π/4+6kπ
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Answers & Comments
2cosx=-1
cosx=-1/2
x=+-arcos a+2kπ, k∈Z
x=+- 2π/3+2kπ
б) tg2x+1=0
tg2x=-1
2x=-π/4+2kπ
x=-π/8+kπ
в)sin(x/3+π/4)=-1
x/3+π/4=-π/2+2kπ
x/3=-π/2+π/4+2kπ
x/3=-π/4+2kπ
x=-3π/4+6kπ