SOLUTION: The answer must be given In radians. Thanks! 2cos^2x + 3cosx -2 = 0

Question 1045447: The answer must be given In radians. Thanks!
2cos^2x + 3cosx -2 = 0
Found 2 solutions by stanbon, ikleyn:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The answer must be given In radians. Thanks!
2cos^2x + 3cosx -2 = 0
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Factor::
(2cos(x)-1)(cos(x)+2) = 0
------
cos(x) = 1/2 or cos(x) = -2 (extraneous)
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x = arccos(1/2) = pi/3 or x = (11/3)pi
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Cheers,
Stan H.
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Answer by ikleyn(52887) (Show Source): You can put this solution on YOUR website!
.
The answer must be given In radians. Thanks!
2cos^2x + 3cosx -2 = 0
----
Factor::
(2cos(x)-1)(cos(x)+2) = 0
------
cos(x) = 1/2 or cos(x) = -2 (extraneous)
-----
x = arccos(1/2) = pi/3 or x = (5/3)pi

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