Find the sum of all of the solutions to the equation cos(2x) - sin2(x) - 1 = 0 in the interval 0 < x < 5 pi. (Use radian measure for x.) (A) 15pi (B) 0 (C) pi (D) 10pi (E) None of the above.
cos(2x)-1=-2sin^2(x)
sin(2x)=2sin(x)cos(x)
The equation is
-2sin(x)(sin(x)+cos(x))=0
sin(x)=0 or sin(x)+cos(x)=0
sin(x)=0==>x=pi,2pi,3pi,4pi
sin(x)+cos(x)=0,
sin(x)=-c0s(x)
tan(x)=-1==>x=3pi/4,7pi/4,11pi/4,15pi/4,
19pi/4
(E)
To find the sum of all the solutions to the given equation, we first need to solve the equation in the given interval (0 < x < 5 pi).
The equation cos(2x) - sin^2(x) - 1 = 0 can be simplified using the trigonometric identity: sin^2(x) = 1 - cos^2(x).
Substituting this identity into the equation, we get:
cos(2x) - (1 - cos^2(x)) - 1 = 0
Rearranging and simplifying, we have:
cos(2x) + cos^2(x) - 2 = 0
To solve this equation, we can use the quadratic formula:
cos^2(x) + cos(2x) - 2 = 0
Let's denote cos(x) = t. Substituting, we have:
t^2 + 2t - 2 = 0
Using the quadratic formula, we find the possible values of t:
t = (-2 ± √(2^2 - 4 * 1 * -2)) / 2
t = (-2 ± √(4 + 8)) / 2
t = (-2 ± √12) / 2
t = (-2 ± 2√3) / 2
t = -1 ± √3
Since cos(x) = t, we can now solve for x:
cos(x) = -1 ± √3
To find the solutions in the given interval (0 < x < 5 pi), we need to consider values of x for which 0 < x < 5 pi and cos(x) = -1 ± √3.
The only solution in the given interval is x = 2pi/3.
Therefore, the sum of all solutions to the equation in the given interval is x = 2pi/3.
The answer is (C) pi.