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Kolyan22812
@Kolyan22812
July 2022
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помогите решить тригонометрическое уравнение 5соs(2x+п/2)=2,5
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Amigo3
5соs(2x+π/2)=2,5⇒cos(2x+π/2)=0,5⇒2*x1+π/2=π/3+2*π*N⇒
x1=(π/3+2*π*N-0,5*π)/2=π*N-0,25*π+π/6.
2*x2+π/2=5*π/3+2*π*N⇒х2=π*N-0,25*π+2,5*π/3.
Ответ: x1=π*N-0,25*π+π/6;
х2=π*N-0,25*π+2,5*π/3, N
∈Z.
1 votes
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Answers & Comments
x1=(π/3+2*π*N-0,5*π)/2=π*N-0,25*π+π/6.
2*x2+π/2=5*π/3+2*π*N⇒х2=π*N-0,25*π+2,5*π/3.
Ответ: x1=π*N-0,25*π+π/6; х2=π*N-0,25*π+2,5*π/3, N∈Z.