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Trig equation, need help

  1. Aug 21, 2013 #1
    1. The problem statement, all variables and given/known data

    tanx-1 = cos2x

    3. The attempt at a solution

    I know tanx = sinx/cosx but I don't know which identity to pick for cos2x
     
  2. jcsd
  3. Aug 21, 2013 #2

    Mark44

    Staff: Mentor

    Pick one and see if it works.
     
  4. Aug 21, 2013 #3
    Good day TheRedDevil18

    If you replace cos(2x) with 2cos2(x) - 1 ,
    that will get rid of the (-1) on the left side. Then go from there.......
     
  5. Aug 22, 2013 #4
    sinx/cosx = 2cos^2(x)
    sinx/cosx = 2(1-sin^2(x))
    sinx/2cosx + sin^2(x) = 1
    sinx/2cosx * 1/sin^2(x) = 1
    2SinxCosx = 1
    sin2x = 1
    x = 45

    All good?
     
  6. Aug 22, 2013 #5

    Mark44

    Staff: Mentor

    ???
    What are you doing here (above)?
    No.
    If you check your work, you'll see that 45° is not a solution.

    In addition, there probably an infinite number of solutions, unless the problem specifies that you should find only the solutions in a certain interval. Since that's not stated in your post, you need to list all the solutions.
     
  7. Aug 22, 2013 #6
    45° satisfies the given equation but RedDevil reached it with a wrong method as you pointed out.
     
  8. Aug 22, 2013 #7

    Mark44

    Staff: Mentor

    Thanks! He didn't show the original equation in his later work, and I misremembered cos2x as cos2x in my check.
     
  9. Aug 22, 2013 #8

    verty

    User Avatar
    Homework Helper

    I can't see how to solve this in an easy way. I get to

    ##sin(x) = 2 \; cos^3(x)##

    and am stuck. I suspect something is wrong with this question.
     
  10. Aug 22, 2013 #9
    Remember to specify interval or say "45°+ 360°(n)" as Mark44 pointed out
     
  11. Aug 22, 2013 #10

    haruspex

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    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Go back to ##tan(x) = 2 \; cos^2(x)## and write ##cos^2## as ##1/sec^2 = 1/(1+tan^2)## to get a cubic in tan(x). One solution is obvious, the other two are complex.

    RD18 turned sinx/2cosx + sin2(x) = 1 into sinx/2cosx * 1/sin2(x) = 1 instead of sinx/2cosx * 1/sin2x +1 = 1/sin2x, but got the right answer through sheer luck.
     
  12. Aug 23, 2013 #11

    ehild

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    Homework Helper
    Gold Member

    cos2x=1/(1+tan2x)

    So you can write the equation as

    tan3x+tanx-2=0,

    rearrange: (tan3x-1)+(tanx-1)=0

    factorize: The common factor is tanx-1.

    ehild
     
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