text: I’d suggest you another approach: (cosx)^2=0.5*(1+cos(2x)); if y’=0 then 1+cos(2x)=0.4*x or cos(2x)=0.4*x-1 (the x=0 is outside of our interval). Now look at the graphs of y1=cos(2x) and that of y2=0.4x-1; |cos(2x)|<=1; ‘start’ point of  y2 is [0,-1], y2 is rather ‘slow-up’ and y2 must be <= than 1, ie 0.4*x-1<=1 or x<=5; thus our interval is not (0,10) any longer, it is (0,5]; \n5=1.59*pi; 1.59 is the # of periods of y1 in 5, because 1 period of y1 is 2x=2*pi and x=pi.\nThus there are only 3 critical points in the interval (0,10)!