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Trigonometric Equation solving

  1. May 31, 2013 #1
    1. The problem statement, all variables and given/known data

    cos 2x + cos 4x = cos x

    0 degrees ≤ X ≤ 360 degrees

    2. Relevant equations
    Trig identities.


    3. The attempt at a solution

    Trig identities and equations are my weakness so I'm trying to review some to get better at them, so I guess this isn't technically homework. But it's out of my textbook.

    Can someone give me a kick for the right start?

    I tried:

    [tex]2cos(\frac{(2x + 4x)}{2}) cos(\frac{(2x-4x)}{2})[/tex]

    [tex]2cos3x * cosx = cosx[/tex]

    [tex] 2cos3x = 1 [/tex]

    [tex]cos3x = 1/2[/tex]

    I'm not sure if this is right or where to go from here.
     
  2. jcsd
  3. May 31, 2013 #2

    Mark44

    Staff: Mentor

    As stated, it's not clear what you need to do. Are you supposed to solve an equation or prove an identity? From your work below, you are apparently solving an equation.
    How is this related to cos(2x) + cos(4x)? It could be that the above equals cos(2x) + cos(4x), but this is not an identity that I have committed to memory.
    Assuming for the moment that your work is correct, you should never do what you just did, since you lost a bunch of solutions. You pretty much never want to divide by a variable expression, because it's possible that expression might be zero for some value(s) of the variable.

    Instead, bring everything over to one side, and then factor it.
    Can you solve cos(b) = 1/2?
     
  4. May 31, 2013 #3

    CAF123

    User Avatar
    Gold Member



    On line *, where you 'divide out' by cosx, you are implicitly assuming cosx is not zero here, that is that x is not ##\pi/2 + n\pi##. Do you think this is wise?
     
  5. May 31, 2013 #4
    Trying to solve the equation to find the angle.

    It is related to cos(2x) + cos(4x) because those are the identities for addition of cos.

    cos u + cos b = 2cos((u+v)/2) * cos((u-v)/2)

    So I guess what I did was wrong then? So I have no idea where to go, unfortunately.

    I just find everything in trig so far to be confusing, I was just starting to understand using the identities and so on, but coming to this, just confuses me and I can't get around the mental block of it.
     
  6. May 31, 2013 #5

    Mark44

    Staff: Mentor

    I didn't say that. The identity you used is fine, I believe. It's just not one that I've used a lot.


    You got to here:
    2cos(3x)cos(x) = cos(x)

    Bring everything over to one side and then factor what you have.
     
  7. May 31, 2013 #6
    2cos(3x)cos(x) - cos(x) = 0

    cos(x) ( 2cos(3x) - 1) = 0

    so cos(x) = 0
    x = 90 , x = 180


    2cos(3x) - 1 = 0

    2(cos(3x) = 1

    cos(3x) = 1/2

    IS this right so far?
     
  8. May 31, 2013 #7

    Mark44

    Staff: Mentor

    Mostly. cos(x) = 0 ==> x = 90° but does not imply that x = 180°. cos(180°) = -1.

    Everything else looks fine.

    A question I asked earlier was this: Can you solve cos(b) = 1/2?
     
  9. May 31, 2013 #8
    Oops, wasn't thinking when I put 180. I meant 270.

    cos(b) = 1/2
    b = 60 degrees



    cos 3x = 1/2

    cos x = 60

    3x = 60

    x = 20


    So it checks correct when I sub it back in the equation. Did I get everything?
     
    Last edited: May 31, 2013
  10. Jun 1, 2013 #9

    Mark44

    Staff: Mentor

    Actually, b = 60° + k * 360°, where k is an integer
    3x = 60° + k * 360°, where k is an integer
    So x = ?

    You want values of x such that 0 ≤ x ≤ 360°
    See above.
     
  11. Jun 1, 2013 #10
    3x = 60° + k * 360°, where k is an integer
    So x = ?

    would it be 20° + k * 360°?
     
  12. Jun 2, 2013 #11

    Mark44

    Staff: Mentor

    No.
    You're multiplying both sides of the equation 3x = 60° + k * 360°, but you forgot to distribute the 1/3 to both terms.
     
  13. Jun 2, 2013 #12

    x = 20 + (360k)/3
     
  14. Jun 2, 2013 #13

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Yep. And (360k)/3 = ...k ?
     
  15. Jun 2, 2013 #14
    120k
     
  16. Jun 2, 2013 #15

    Mark44

    Staff: Mentor

    Yes, but you're not quite finished.
    So x = 20° + k * 120°, for k an integer.

    In the problem statement, 0° ≤ x ≤ 360°, so what values of k give x values in this range?
     
  17. Jun 2, 2013 #16
    I think you will be missing solutions. How will you get x=100°,220°, or 340° from this? cos(3x)=1/2 does not mean that 3x=60°+360°k for integer k.
     
    Last edited: Jun 2, 2013
  18. Jun 2, 2013 #17
    20 + 1 * 120 = 140

    20 + 2 * 120 = 260
     
  19. Jun 2, 2013 #18
    0 is an integer too!!! :tongue:

    So, what is our complete solution set? Id est,

    $$\cos{2x}+\cos{4x}=\cos{x} \\ (0^o \leq x \leq 360^o) \\ \forall x\in\{?\}$$
     
  20. Jun 2, 2013 #19

    Our answers should be: 90°, 270°, 20°, 140°, 260°
     
  21. Jun 2, 2013 #20
    Reread my post #16. There are other solutions you can get from saying 3x=300+360k. You will see that x=100°,220°, and 340° will also satisfy your equation.
     
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