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Homework Help: Urgent: Solve for x in this trigonometric equation?

  1. Jan 5, 2010 #1
    1. The problem statement, all variables and given/known data
    Solve for x. x is greater than or equal to 0 and x is less than or equal to 2pi.

    (4/3)sinxcosx-sin22x=0

    If there is a better way to approach this question than the way I did it below, please post it.
    Thanks.


    2. Relevant equations

    N/A

    3. The attempt at a solution

    (4/3)sinxcosx-sin22x=0
    2(2sinxcosx)-3sin22x=0
    2(sin2x)-3sin22x=0
    2sin2x-3sin22x=0
    sin2x(2-3sinx)=0 <------I know this is incorrect. How do I factor the previous line correctly?

    To solve the first part:
    sin2x=0
    2x=sin-1(0)=0,pi,2pi,3pi,4pi
    therefore x=0, pi/2, pi, 3pi/2, 2pi

    I know there are other solutions for x. Please help!
     
    Last edited: Jan 5, 2010
  2. jcsd
  3. Jan 5, 2010 #2

    ehild

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    You just left out the factor "2" in front of x : the second factor is 2-3sin(2x).

    As you have found out, the first set of solutions is x= k*pi/2, with k integer.
    You get the other set by equating the second factor to zero: 2-3sin(2x)=0

    ehild
     
  4. Jan 5, 2010 #3
    Looks like you can solve it using the substitution [STRIKE]u = tan x[/STRIKE] u = tan(x/2)
     
    Last edited: Jan 6, 2010
  5. Jan 5, 2010 #4
    How do I do that? Can you please explain it more clearly? Thanks.
     
  6. Jan 5, 2010 #5
    So if I solve for x in 2-3sin(2x)=0, I'll get the following in rad: 0.365, 1.205, 3.505, 4.345?
    Someone..please correct me if I am wrong.
     
  7. Jan 5, 2010 #6

    ehild

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    Your numbers are correct but you have infinite number of solutions, it is not enough to write down some of them.

    2x=0.730+k*2pi , or
    2x=pi-0.730+2k*pi=2.41+2k*pi, with k an integer (positive or negative or zero).

    So your solutions are in general form:

    x=0.365 + k*pi,

    x=1.206 + k*pi,

    and x=k*pi/2 ,

    with k = integer.

    k can be negative, so x can be also -2.776 or -1.936 ...

    ehild
     
  8. Jan 6, 2010 #7
    I had a mistake in my last message which is corrected now.

    Using the substitution u = tan(x/2) gives us sin x = 2u/(u2 + 1) and cos x = (1 - u2)/(u2 + 1). Substitute those into your equation and you can rewrite it into a quadratic equation in terms of u. See if you can solve for x after you substitute tan(x/2).
     
  9. Jan 6, 2010 #8

    vela

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    The problem statement restricts x to the interval [tex][0,2\pi][/tex].
     
  10. Jan 6, 2010 #9

    ehild

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    Ok, your solution is correct then.

    ehild
     
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