Question 941795
{{{ cos(2x) =2cos^2(x)-1}}}
    3cos2x+cosx+1=0
  {{{ 3(2cos^2(x)-1)+cosx+1=0}}}
 {{{  6cos^2(x) -3 +cosx+1=0}}}
 {{{ 6cos^2(x)+cos(x) -2=0}}}
let assume cos(x) =t
 the above equation becomes as follows
 {{{ 6t^2+t-2=0}}}
 {{{6t^2+4t-3t-2=0}}}
 {{{2t(3t+2)-1(3t+2)=0}}}
 {{{ (3t+2)(2t-1)=0}}}
  possible solotutions
 either (3t+2)=0         or (2t-1)=0
           3t=-2               2t =1  
          t= -2/3                  t= 1/2
Hence t= -2/3 or 1/2
  but t= cos(x) 
 hence cos(x) = -2/3 or 1/2
    x= 131.81 degrees or 60 degrees