
#1
Mar1611, 12:17 PM

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 



#2
Mar1611, 12:27 PM

Mentor
P: 21,012

Next, replace cos(2x) with 2cos^{2}  1. This produces an equation that is quadratic in form. 



#3
Mar1611, 01:14 PM

P: 697

There is a much easier way.




#4
Mar1611, 01:28 PM

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? 



#5
Mar1611, 03:31 PM

P: 697

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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