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August 30, 2014

August 30, 2014

Posted by **Abbey** on Sunday, February 5, 2012 at 11:08pm.

Cosx-cos2x

- calculus -
**Reiny**, Monday, February 6, 2012 at 12:45amI will assume your "equation" is

cosx - cos 2x = 0

cosx - (2cos^2 x - 1) = 0

2cos^2 x - cosx - 1 = 0

(2cosx + 1)(cosx - 1) = 0

cosx = -1/2 or cosx = 1

if cosx = -1/2 , then

x = 2π/3 or x = 4π/3 , (120° or 240°)

if cosx = 1 , then

x = 0 or 2π , (0° or 360°)

- calculus -
**Abbey**, Monday, February 6, 2012 at 8:12amThanks!

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