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justloyrens
@justloyrens
July 2022
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Туплю ужасно.
Если перед функцией -cos(6x-pi/4)=0 стоит минус, то как быть?
6x-pi/4=( минус косинус нуля или как?) pi/2 + pi n,потом прибавляю pi/4 и делю на 6? ИЛИ пишу -pi/2+pi n, потом делаю те же действия.
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Vasily1975
Verified answer
-cos(6*x-π/4)=0,
cos(6*x-π/4)=0,
6*x-π/4=π*(2*n+1)/2=π*n+π/2,
6*x=π*n+π/2+π/4=π*n+3*π/4,
x=π*n/6+π/8
Ответ: x=π*n/6+π/8, n∈Z.
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Answers & Comments
Verified answer
-cos(6*x-π/4)=0,cos(6*x-π/4)=0,
6*x-π/4=π*(2*n+1)/2=π*n+π/2,
6*x=π*n+π/2+π/4=π*n+3*π/4,
x=π*n/6+π/8
Ответ: x=π*n/6+π/8, n∈Z.