Finding sum of roots of trigonometric equation

[Priyadarshi Raj]

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.

 

[Fightfish]

The trick here, I believe, is to realize that reduction of the $33x$ 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 $\sec^2 33x = 1 + \tan^2 33x$ and keep the $\cos 2x$ term as it is.

Edit: I realized a much much simpler method. Note that the LHS is $\geq 0$. How about the RHS? This gives you the answer rightaway.

 

[Ray Vickson]

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?

 

[Priyadarshi Raj]

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.