`cos(2x)-cos(x)=0 , 0<=x<=2pi`

using the identity `cos(2x)=2cos^2(x)-1`

`cos(2x)-cos(x)=0`

`2cos^2(x)-1-cos(x)=0`

Let cos(x)=y,

`2y^2-y-1=0`

solving using the quadratic formula,

`y=(1+-sqrt((-1)^2-4*2(-1)))/(2*2)`

`y=(1+-sqrt(9))/4=(1+-3)/4=1,-1/2`

`:. cos(x)=1, cos(x)=-1/2`

cos(x)=-1/2

General solutions are,

`x=(2pi)/3+2pin, x=(4pi)/3+2pin`

Solutions for the range `0<=x<=2pi` are,

`x=(2pi)/3 , x=(4pi)/3`

cos(x)=1

General solutions are,

`x=0+2pin`

solutions for the range `0<=x<=2pi` are,

`x=0 , x=2pi`

combine all the solutions ,

`x=0, x=2pi , x=(2pi)/3 , x=(4pi)/3`

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