Finding sum of roots of trigonometric equation

  • #1

Homework Statement


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

Homework Equations


## cos(2x) = 2cos^2(x) -1 ##

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.
 

Answers and Replies

  • #2
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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.
 
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  • #3
Ray Vickson
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Homework Statement


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

Homework Equations


## cos(2x) = 2cos^2(x) -1 ##

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.
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?
 
  • #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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