Find the sum of all of the solutions to the equation cos(2x) − sin2(x) − 1 = 0 in theinterval 0 < x < 5 pi. (Use radian measure for x.)

(A) 15pi (B) 0 (C) pi (D) 10pi (E) None of the above.

To find the sum of all the solutions to the equation cos(2x) - sin^2(x) - 1 = 0 in the interval 0 < x < 5pi, we first need to solve the equation in order to find the values of x that satisfy it.

We start by rewriting the equation: cos(2x) - (1 - cos^2(x)) - 1 = 0
Now, simplify the equation: cos(2x) - 1 + cos^2(x) - 1 = 0
Combine like terms: cos^2(x) + cos(2x) - 2 = 0

To solve this equation, we can use the fact that cos(2x) = 2cos^2(x) - 1. By substituting this into the equation, we get:
cos^2(x) + 2cos^2(x) - 1 - 2 = 0
3cos^2(x) - 3 = 0

Divide the entire equation by 3:
cos^2(x) - 1 = 0

Now, factor the equation:
(cos(x) - 1)(cos(x) + 1) = 0

This gives us two possible solutions:
cos(x) - 1 = 0 --> cos(x) = 1
cos(x) + 1 = 0 --> cos(x) = -1

For the first equation, cos(x) = 1, we know that this is true when x = 0. However, the interval given is 0 < x < 5pi, so x = 0 lies in the interval.

For the second equation, cos(x) = -1, we know that this is true when x = pi. Again, we check if pi is within the given interval, but it is not.

Therefore, there are no solutions to the equation cos(2x) - sin^2(x) - 1 = 0 in the interval 0 < x < 5pi.

The correct answer is option (E) None of the above.