Precalculus, Chapter 5, 5.5, Section 5.5, Problem 9

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