SOLUTION: solve to the nearest degree for all values on x in the interval 0≤x≤360, 2cos^2x-4cosx+1=0

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Question 865997: solve to the nearest degree for all values on x in the interval 0≤x≤360, 2cos^2x-4cosx+1=0
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Let u=cos%28x%29
2u%5E2-4u%2B1=0
2%28u%5E2-2u%29%2B1=0
2%28u%5E2-2u%2B1%29%2B1-2=0
2%28u-1%29%5E2=1
%28u-1%29%5E2=1%2F2
u-1=0+%2B-+sqrt%282%29%2F2
u=1+%2B-+sqrt%282%29%2F2
cos%28x%29=1+%2B-+sqrt%282%29%2F2
Since abs%28cos%28x%29%29%3C=1, throw out the positive value.
cos%28x%29=1-sqrt%282%29%2F2
x=cos%5E%28-1%29%281-sqrt%282%29%2F2%29%29
x=72.9 and x=287.1