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Mathematics 20 Online
OpenStudy (anonymous):

Solve for x 2cos^2 (1/2 x) - 2 = 2 cosx

OpenStudy (anonymous):

this equation \[2\cos^2 (\frac{ 1 }{ 2x }) - 2 = 2 cosx\]

OpenStudy (anonymous):

its x/2

OpenStudy (anonymous):

Use half-angle identity: cos^2(a)=1/2[1+cos(2a)] Rearrange into: 2cos^2(a)-1=cos(2a) where a =x/2 Should work...

OpenStudy (anonymous):

Let me try it :)

OpenStudy (anonymous):

I'm kind of confused on how you got the 1/2[1+cos(2a)] part. I know you substituted 2a for x but idk how you got the 1/2

OpenStudy (dan815):

that is the formula

OpenStudy (anonymous):

Oh, yea xD

OpenStudy (anonymous):

You are asked to solve for x? Because the solution is cyclic. Basically your question is the half-angle identity and anything in the range: 0 < x < pi over 2 will satisfy it.

OpenStudy (anonymous):

I'm only happy with a whole pie though. ;)

OpenStudy (anonymous):

mmmmmm cherry

OpenStudy (dan815):

do this

OpenStudy (dan815):

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