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dsocol
@dsocol
July 2022
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Решить уравнение[tex]sin ^{2} x=1[/tex]
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Munkush
Sin^2x =1
sinx = 1 , -1
sinx=1
x=arcsin1
x=п/2
x=п-п/2
Период 2п
x=п/2 ± 2пn ; п/2 ± 2пn
sinx=-1
x=arcsin-1
x=-п/2
х=2п+п/2+п
x=3п
Период 2п
х=-п/2±2пn ; 3пn/2 ± 2пn
х=3п/2 ± 2пn
ответ : x=п/2 ± 2пn ; 3п/2± 2пn
1 votes
Thanks 1
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Answers & Comments
sinx = 1 , -1
sinx=1
x=arcsin1
x=п/2
x=п-п/2
Период 2п
x=п/2 ± 2пn ; п/2 ± 2пn
sinx=-1
x=arcsin-1
x=-п/2
х=2п+п/2+п
x=3п
Период 2п
х=-п/2±2пn ; 3пn/2 ± 2пn
х=3п/2 ± 2пn
ответ : x=п/2 ± 2пn ; 3п/2± 2пn