# How to solve for x in this trig jumble?

by zeion
Tags: jumble, solve, trig
 P: 467 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
Mentor
P: 19,704
 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.
 P: 697 There is a much easier way.
P: 63

## 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?
P: 697
 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

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