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

Question 855999: Solve for all real values of x in radians. AMP Parsing Error of [2cos^2x-5cosx+2=0]: Invalid function '\x-5\cosx+2=0': opening bracket expected at /home/ichudov/project_locations/algebra.com/templates/Algebra/Expression.pm line 70. .
Found 2 solutions by Alan3354, KMST:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Solve for all real values of x in radians.
(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) (Show Source): You can put this solution on YOUR website!
meaning
is easier to solve by thinking in terms of
and seeing the equation as

The solutions to that equation (no matter how you find them) are
and
Going back to , has no solution,
but has an infinite number of solutions.
In the first quadrant, ,
and in the fourth quadrant .
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
or maybe even .
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