SOLUTION: Solve for x where 0 degrees is less then or equal to x and x is less then or equal too 360 degrees 6cos^2x-5cosx-6=0

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Question 1013578: Solve for x where 0 degrees is less then or equal to x and x is less then or equal too 360 degrees
6cos^2x-5cosx-6=0

Found 2 solutions by stanbon, MathTherapy:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x where 0 <= x <= 360 degrees
6cos^2x-5cosx-6 = 0
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(6cos(x)+1)(cos(x)-1) = 0
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cos(x) = -1/6 or cos(x) = 1
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x = arccos(-1/6) = 99.59 degrees or 360-99.59 = 260.41 degrees
OR
x = arccos(1) = 0 degrees
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Cheers,
Stan H.
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Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x where 0 degrees is less then or equal to x and x is less then or equal too 360 degrees
6cos^2x-5cosx-6=0

OR
cos+%28x%29+=+3%2F2(ignore)