Question 399928
{{{2(cosw)^2+2=5cosw}}}
{{{2(cosw)^2-5cosw+2=0}}}
make the substitution {{{x=cosw}}}, x have to be from the interval [-1;1](because cosw always takes values only in the interval[-1;1])
{{{2x^2-5x+2=0}}}
 {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
 {{{x = (-(-5) + sqrt( (-5)^2-4*2*2 ))/(2*2) }}} or  {{{x = (-(-5) - sqrt( (-5)^2-4*2*2 ))/(2*2) }}} 
 {{{x = (5 + sqrt( 9))/4 }}} or  {{{x = (5 - sqrt( 9))/4 }}}
 {{{x = 2}}} or  {{{x = 1/2}}}
 {{{x = 2}}} is extraneous root of the equation because it does not belong to the interval [-1;1]
So  {{{x = 1/2}}}, 
make the substitution back {{{x=cosw}}}, {{{cosw=1/2}}}
FORMULA: the solution of eguation {{{cosw=a}}} is w=+-{{{arccos(a)+360*k}}} k is any integer
w=+-{{{arccos(1/2)+360*k}}} k is any integer
w=+-{{{60+360*k}}} k is any integer