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янкарпо
@янкарпо
July 2022
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2 × √3:2sin3x-1:2cos3x=1
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sedinalana
Verified answer
Пользователь написал,что есть скобки
2(√3/2*sin3x-1/2*cos3x)=1
sin3x*cosπ/6-cos3x*sinπ/6=1/2
sin(3x-π/6)=1/2
3x-π/6=π/6+2πn,n∈z U 3x-π/6=5π/6+2πk,k∈z
3x=π/3+2πn,n∈z U 3x=π+2πk,k∈z
x=π/9+2πn/3,n∈z U x=π/3+2πk,k∈z
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Answers & Comments
Verified answer
Пользователь написал,что есть скобки2(√3/2*sin3x-1/2*cos3x)=1
sin3x*cosπ/6-cos3x*sinπ/6=1/2
sin(3x-π/6)=1/2
3x-π/6=π/6+2πn,n∈z U 3x-π/6=5π/6+2πk,k∈z
3x=π/3+2πn,n∈z U 3x=π+2πk,k∈z
x=π/9+2πn/3,n∈z U x=π/3+2πk,k∈z