cos(5x/2)cos(3x/2)+sin(7x/2)sin(x/2)+cos2x=0

Đáp án:

`cos(5x/2)cos(3x/2)+sin(7x/2)sin(x/2)+cos2x=0`

`2cos(5x/2)cos(3x/2)+2sin(7x/2)sin(x/2)+2cos2x=0`

`Cos4x+Cosx +Cos3x-Cos4x+2Cos2x=0`

`=>Cos3x+Cos2x+Cos2x +Cosx=0`

`=>2Cos(5/2*x)Cos(x/2)+2Cos(3/2x)Cos(x/2)=0`

`=>2Cos(x/2)(Cos(5/2x)+Cos(3/2x))=0`

`=>Cos(x/2)=0`

`=>x/2=pi/2+kpi`

`=>x=pi+2kpi`

`Cos(5/2x)=-Cos(3/2x)`

`Cos(5/2x) =Cos( pi+3/2x )`

`=> 5/2x =pi+3/2x+k2pi`

`=> x=pi+k2pi`

hoặc `5/2x =-pi-3/2x+k2pi`

`=>4x=-pi+k2pi`

`=>x=-pi/4+k2pi/2`

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