Why is the answer for solving cos2x=cosx not 0°+360k°?

[WillyTech]

ok here's a question i can't figure out why the answer is the way it is.

Solve each equation for all values of x.

1.cos2x=cosx

*First i use the double identity 2cos²x-1 for cos2x and got:
2cos²x-1=cosx

*then i subtracted cosx and got:
2cos²x-cosx-1=0

*then i subtract 1 on both side and got:
2cos²x-cosx=1

*I then factor 2cos²x-cosx and got:
cosx(2cosx-1)=1

*then i set them equal to 1 and got this:

cosx=1
=0°

2cosx-1=1
added one and divided by 2 to get:
cosx=1
=0°


Since it is cosine i set it like this: 0°+360k°

But my book says the correct answer is 0°+120k°

Please someone help me with this.
And i may have more questions on some other double identity problems.

 

[George Jones]

Why do both factors have to equal 1? For example, (-3)*(-1/3) = 1. Note, I'm not saying these numbers are useful for your question, I'm just pointing out other possibilities.

Try factoring your equation before moving 1 to the RHS.

Regards,
George

 

[VietDao29]

ok here's a question i can't figure out why the answer is the way it is.

Solve each equation for all values of x.

1.cos2x=cosx

*First i use the double identity 2cos²x-1 for cos2x and got:
2cos²x-1=cosx

*then i subtracted cosx and got:
2cos²x-cosx-1=0

So far so good. From here, it's just the quadratic equation in cos x, i.e if you substitute t = cos x, your equation will become:
2t² - t + 1 = 0.
Can you solve for cos x (or t?) from the equation? From there, I think you'll be able to solve for x. Can you go from here? :)

 

[WillyTech]

Why do both factors have to equal 1? For example, (-3)*(-1/3) = 1. Note, I'm not saying these numbers are useful for your question, I'm just pointing out other possibilities.

Try factoring your equation before moving 1 to the RHS.

Regards,
George

So that means i mest up somewhere in the factoring? and its not suppose to both equal 1?

Well, i get this after i factor 2cos²x-cosx-1=0 :

cosx(2cos-1)-1=0

I am really stuck on this problem.

 

[George Jones]

Factor the trinomial into a product of binomials.

Big hint: substitute y = cosx.

Regards,
George

 

[WillyTech]

Factor the trinomial into a product of binomials.

Big hint: substitute y = cosx.

Regards,
George

so it would be: 2y^2-y+1=0

and then would the product of the binomials would be (2y+1)(y-1)?

 

[George Jones]

Right, and the product equals zero. What does this tell you?

Regards,
George

 

[WillyTech]

Right, and the product equals zero. What does this tell you?

Regards,
George

Oh yeah! i see now!

it will come out to cosx=-1/2 and 1 which is "120°" and "0°" which is why it has to be 0°+120k°.

It was just some simple wrong factoring that got me on the wrong track.
Thanks!

 

[George Jones]

Good!

Sorry VietDao29 - I didn't see your post.

Regards,
George

 

[0rthodontist]

Just to illustrate why the terms of your earlier product don't have to equal 1: (1/3)*3 = 1 but neither is 1.