solve 2 - 2sin^2x = cosx over interval pie <= x <= 2pi

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hollerback1

Junior Member
Joined
Dec 21, 2005
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Hello, this problem is bugging me because I can't figure out why one part is factored.
Solve in pie <= x <= 2pie:

2 - 2sin^2x = cosx

so...2 - 2(1-2cos^2x) = cosx

then 2 - 2 + 2cos^2x = cosx?

I'm not sure how this part turned out like this.

Thanks, and Merry Xmas.
 
Re: Identity Problem

hollerback1 said:
Hello, this problem is bugging me because I can't figure out why one part is factored.
Solve in pie <= x <= 2pie:

2 - 2sin^2x = cosx

so...2 - 2(1-2cos^2x) = cosx
They meant to substitute sin^2(x) = 1 - cos^2(x), not 1 - 2cos^2(x).

then 2 - 2 + 2cos^2x = cosx?
This follows from above with the correction.

I'm not sure how this part turned out like this.
Click to expand...
 
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