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Finding sum of roots of trigonometric equation

  1. Apr 19, 2016 #1
    1. The problem statement, all variables and given/known data
    Question:
    Sum of all the solutions of the equation: ##tan^2 (33x) = cos(2x)-1## which lie in the interval ## [0, 314] ## is:
    (a) 5050 π
    (b) 4950 π
    (c) 5151 π
    (d) none of these

    The correct answer is: (b) 4950 π

    2. Relevant equations
    ## cos(2x) = 2cos^2(x) -1 ##

    3. The attempt at a solution
    ##tan^2 (33x) = cos(2x)-1##
    ⇒ ##tan^2 (33x) = 2cos^2(x)-2##
    ⇒ ##\frac{sin^2 (33x)}{cos^2(33x)} = 2(cos^2 (x) -1)##
    ⇒ ##sin^2 (33x) = 2cos^2(x)cos^2(33x) - 2cos^2 (33x)##
    ⇒ ##1 - cos^2(33x)= 2cos^2(x)cos^2(33x) - 2cos^2 (33x)##
    ⇒ ##2cos^2(x)cos^2(33x) - cos^2 (33x) - 1 = 0##

    And I end up making it more complex. It'b be great if I could convert the ##33x## to something smaller.

    Please help.
     
  2. jcsd
  3. Apr 19, 2016 #2
    The trick here, I believe, is to realise that reduction of the [itex]33x[/itex] is probably not a feasible option at all - so what you want to get in the end is not a quadratic or polynomial equation, but a simple product of trigonometric functions being equal to a nice number.

    So the hint is to use the trigonometric identity [itex]\sec^2 33x = 1 + \tan^2 33x[/itex] and keep the [itex]\cos 2x[/itex] term as it is.

    Edit: I realised a much much simpler method. Note that the LHS is [itex]\geq 0[/itex]. How about the RHS? This gives you the answer rightaway.
     
  4. Apr 19, 2016 #3

    Ray Vickson

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    Since ##\cos(2x)-1 = \cos^2(x) - \sin^2(x)-1 = 1-2 \sin^2(x)-1 = -2 \sin^2(x)##, your equation becomes ##\tan^2(33x) + 2 \sin^2(x) = 0##. When can a sum of squares equal zero?
     
  5. Apr 22, 2016 #4
    In both the ways I get the solutions as: ## \pi, 2\pi, 3\pi ... ##

    So the required sum is ## = \pi + 2\pi + 3\pi +... + 99\pi = 4950\pi ##

    Thanks everyone.
     
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