How to solve for x in this trig jumble?

1. The problem statement, all variables and given/known data

Basically I need to solve for x that satisfies

2cosx - 2cos2x = 0

2. Relevant equations

3. The attempt at a solution

I changed it to 4sin(x/2)sin(3x/2) = 0
Then I know that sin(0) and sin(pi) = 0
So solve for x(1/2) = pi and x(3/2) = pi.
I get 4pi/3 and the other one is out of the indicated bound.
But this is not the only solution. How do I get all the solutions?

Thanks
 PhysOrg.com science news on PhysOrg.com >> King Richard III found in 'untidy lozenge-shaped grave'>> Google Drive sports new view and scan enhancements>> Researcher admits mistakes in stem cell study

Mentor
 Quote by zeion 1. The problem statement, all variables and given/known data Basically I need to solve for x that satisfies 2cosx - 2cos2x = 0 2. Relevant equations 3. The attempt at a solution I changed it to 4sin(x/2)sin(3x/2) = 0 Then I know that sin(0) and sin(pi) = 0 So solve for x(1/2) = pi and x(3/2) = pi. I get 4pi/3 and the other one is out of the indicated bound. But this is not the only solution. How do I get all the solutions? Thanks
First off, divide both sides by 2.
Next, replace cos(2x) with 2cos2 - 1. This produces an equation that is quadratic in form.
 There is a much easier way.

How to solve for x in this trig jumble?

Do as Mark said, and the other solutions should become apparent. You didn't specify what the bound was, but remember that, for example, if sin(x) = 1/2, (assuming it is unbounded) the solutions will be π/6 +2kπ and 5π/6 + 2kπ . There may be more than one solution for each.

By the way, I'm interested in the easier way to solve it. Can you send me a message giving me a hint? It's killing me, I can't seem to figure out how to do it easier. Or was the easier way the quadratic?

 Quote by mharten1 By the way, I'm interested in the easier way to solve it.
Since the approach already given is pretty simple in itself, I guess there is no problem posting the other way, which seems simpler to me. Either way is pretty straightforward. However, to me it seems more direct to do this.

2cosx - 2cos2x = 0
2cosx = 2cos2x
cosx = cos2x
acos(cosx) = acos(cos2x) with proper consideration for periodicity (-pi<x<pi) and proper domains for acos(arg)

Of course, one can get the answer visually from here, but to be formal consider the proper domains for acos function (i.e. 0<=arg<pi)

for 0<x<pi/2 we get x=2x which gives x=0

for pi/2<x<pi we get x=2(-x+pi) which gives x=2pi/3

then symmetry about x=0 gives x=-2pi/3

So the three solutions in one period are x=0, 2pi/3, -2pi/3

Then these answers will repeat every integer multiple of 2pi