# Finding sum of roots of trigonometric equation

## 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.

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.

Samy_A and member 587159

## 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.