How do you find all values of x in the interval [0, 2pi] in the equation $cos2x= cosx$ ?

Question

3668 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-09T14:26:58

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

Using the identity $cos2x=2cos^2x-1$ , $cos2x=cosx$ becomes

$2cos^2x-1=cosx$ or

$2cos^2x-cosx-1=0$

Using quadratic formula

$cosx=(-(-1)+-sqrt((-1)^2-4xx2xx(-1)))/(2xx2)$ or

$cosx=(1+-sqrt(1+8))/4$ or $cosx=(1+-3)/4$ i.e.

$cosx=1 or -1/2$

Hence $x=0, (2pi)/3, (4pi)/3, 2pi$

How do you find all values of x in the interval [0, 2pi] in the equation cos2x= cosxcos2x= cosx?