сosx=t
2t^2-t-1=0;
D=b^2-4ac=1+8=9;
t1=(1+3)/4 = 4/4 = 1 => cosx = 1 => x=2pi*n, n=0,1,2,3...
t2(1-3)/4=-2/4=-1/2 => cosx = -1/2 => x=+-2pi/3 + 2pi * k, k=0,1,2,3...
Самые большие корни промежутка [0;2pi]:
x=2pi и x=4pi/3
Их сумма равна 2pi + 4pi/3 = 6pi+4pi/3 = 10pi/3