trigonometry
posted by lan .
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)