SOLUTION: Solve for 0&#8804;x<2pi: 1-((sin^2x)/(1-cosx))=(1/2) (Hint: simplify left side with identities)

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Question 1030598: Solve for 0≤x<2pi:
1-((sin^2x)/(1-cosx))=(1/2)
(Hint: simplify left side with identities)
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
1-((sin^2x)/(1-cosx))=(1/2)
1-(1-cos^2x)/(1-cos x)=(1/2)
1- [(1+cos x)(1- cos x)/(1-cos x)]=(1/2)
1-(1+cos x)=(1/2)
1-1-cos x=(1/2)
cos x=(-1/2)
on the 0 to 2 pi interval, that occurs at 2 pi/3 and 4 pi/3.
Where the cosine is -0.5, the sine is +or- sqrt(3)/2
[1-(3/4)/(3/2)]=1-(1/2)=(1/2)