Question 121956
*[Tex \LARGE \cos\left(2x\right)=\cos\left(x\right)] Start with the given equation
 

*[Tex \LARGE 2\cos^{2}\left(x\right)-1=\cos\left(x\right)] Replace *[Tex \LARGE \cos\left(2x\right)] with *[Tex \LARGE 2\cos^{2}\left(x\right)-1]. Note: I'm using the identity *[Tex \LARGE \cos\left(2x\right)=2\cos^{2}\left(x\right)-1]
 

*[Tex \LARGE 2\cos^{2}\left(x\right)-1-\cos\left(x\right)=0] Subtract {{{cos(x)}}} from both sides
 

*[Tex \LARGE 2\cos^{2}\left(x\right)-\cos\left(x\right)-1=0] Rearrange the terms
 

Now let {{{u=cos(x)}}}


*[Tex \LARGE 2u^2-u-1=0] Replace each {{{cos(x)}}} with {{{u}}}
 

{{{(u-1)(2u+1)=0}}} Factor the left side

 
Now set each factor equal to zero


{{{u-1=0}}} or {{{2u+1=0}}}


Now solve for u in each case


{{{u=1}}} or {{{u=-1/2}}}


Now remember we let {{{u=cos(x)}}}. So this means
 

{{{cos(x)=1}}} or {{{cos(x)=-1/2}}}
 

So let's solve {{{cos(x)=1}}} to get {{{x=0}}} or {{{x=2pi}}}. However, since {{{2pi}}} is excluded the only solution for {{{cos(x)=1}}} is {{{x=0}}}
 

Now let's solve {{{cos(x)=-1/2}}} to get {{{x=2pi/3}}} or {{{x=4pi/3}}}
 

So putting these solutions together, we get:


{{{x=0}}}, {{{x=2pi/3}}} or {{{x=4pi/3}}}
 


So this means the answer is B)