Question 855999
{{{2cos^2(x)-5cos(x)+2=0}}} meaning {{{2(cos(x))^2-5cos(x)+2=0}}}
is easier to solve by thinking in terms of
{{{y=cos(x)}}} and seeing the equation as
{{{2y^2-5y+2=0}}}
The solutions to that equation (no matter how you find them) are
{{{y=2}}} and {{{y=1/2}}}
Going back to {{{x}}} , {{{cox(x)=2}}} has no solution,
but {{{cos(x)=1/2}}} has an infinite number of solutions.
In the first quadrant, {{{cos(pi/3)=1/2}}} ,
and in the fourth quadrant {{{cos(-pi/3)=1/2}}} .
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*pi +- pi/3}}} or maybe even {{{(2n+1/3)pi}}} .