Posted by **lan** on Thursday, May 12, 2011 at 7:45pm.

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.

- trigonometry -
**Mgraph**, Thursday, May 12, 2011 at 8:37pm
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)

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