Since cos(2x) = 1 at x = 0,pi,2pi,3pi,...
sec(x) = 1 at 0,2pi,4pi,6pi,...
They meet at even multiples of pi.
cos never gets above 1, sec never gets below 1.
On your interval, x=0 is the only solution
cos 2x = sec x
2cos^2 x - 1 = 1/cosx
2 cos^3 x - cos x - 1 = 0
cosx = 1 is the only real solution.
cos(2theta)= 120/169; 2theta in QIV