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batalygina
@batalygina
July 2022
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Решите, пожалуйста, срочно.
cos4x-cosx*cos3x=0
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mikael2
Verified answer
Cosx*cos3x=0.5(cos2x+cos4x)
cos4x-cos2x/2-cos4x/2=1/2cos4x-cos2x/2=1/2(2cos²2x-1)-1/2 cos2x
cos²2x-1/2cos2x-1/2=0
cos2x=z
3z²-0.5z-1/2=0
6z²-z-1=0
z=1/2 z=-1/3
cos2x=1/2 2x=+-π/3+2πn x=+-π/6+πn
cos2x=-1/3 2x=+-arccos(5/6π)+2πn x=+-1/2*arccos5π/6+πn
n∈Z
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Answers & Comments
Verified answer
Cosx*cos3x=0.5(cos2x+cos4x)cos4x-cos2x/2-cos4x/2=1/2cos4x-cos2x/2=1/2(2cos²2x-1)-1/2 cos2x
cos²2x-1/2cos2x-1/2=0
cos2x=z
3z²-0.5z-1/2=0
6z²-z-1=0
z=1/2 z=-1/3
cos2x=1/2 2x=+-π/3+2πn x=+-π/6+πn
cos2x=-1/3 2x=+-arccos(5/6π)+2πn x=+-1/2*arccos5π/6+πn
n∈Z