SOLUTION: Solve for all real values of x in radians. {{{2cos^2x-5cosx+2=0}}}

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Question 855999: Solve for all real values of x in radians. 2cos%5E2x-5cosx%2B2=0
Found 2 solutions by Alan3354, KMST:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for all real values of x in radians. 2cos%5E2%28x%29-5cos%28x%29%2B2=0
(2cos(x)) - 1)*(cos(2x) - 2) = 0
------
cos(x) = 1/2
x = pi/3 + 2n*pi, n = 0,1,2,3...
x = 5pi/3 + 2n*pi, n = 0,1,2,3...
====================
cos(x) = 1
x = 0 + n*pi, n = 0,1,2,3...

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
2cos%5E2%28x%29-5cos%28x%29%2B2=0 meaning 2%28cos%28x%29%29%5E2-5cos%28x%29%2B2=0
is easier to solve by thinking in terms of
y=cos%28x%29 and seeing the equation as
2y%5E2-5y%2B2=0
The solutions to that equation (no matter how you find them) are
y=2 and y=1%2F2
Going back to x , cox%28x%29=2 has no solution,
but cos%28x%29=1%2F2 has an infinite number of solutions.
In the first quadrant, cos%28pi%2F3%29=1%2F2 ,
and in the fourth quadrant cos%28-pi%2F3%29=1%2F2 .
Adding those solutions plus all the co-terminal angles differing from those by an integer number of turns,
we can express all the solutions as
2n%2Api+%2B-+pi%2F3 or maybe even %282n%2B1%2F3%29pi .