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TheWitcher161
@TheWitcher161
July 2022
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Срочно нужна помощь!!!
Решите
cosx/2 × (cosx/2 - 1)=0
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laimdemon
Cosx = cos(2·(x\2)) = 2cos^2(x\2) - 1
Замена: cos(x\2) = a
1 + 2a^2 - 1 = a
2a^2 = a
1) a=0
cos(x\2)=0
x\2 = pi\2 + pi·k
х=pi + 2·pi·k, k - целое
2)a = 1\2
cos(x\2)=1\2
x\2 = +\-pi\3 + 2pi·n
х=+\-2pi\3 + 4·pi·n, n - целое
Ответ:
х1 = pi + 2·pi·k
х2 = 2pi\3 + 4·pi·n
х3 = - 2pi\3 + 4·pi·n
n,k - целые
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Answers & Comments
Замена: cos(x\2) = a
1 + 2a^2 - 1 = a
2a^2 = a
1) a=0
cos(x\2)=0
x\2 = pi\2 + pi·k
х=pi + 2·pi·k, k - целое
2)a = 1\2
cos(x\2)=1\2
x\2 = +\-pi\3 + 2pi·n
х=+\-2pi\3 + 4·pi·n, n - целое
Ответ:
х1 = pi + 2·pi·k
х2 = 2pi\3 + 4·pi·n
х3 = - 2pi\3 + 4·pi·n
n,k - целые