Solve for the equation for the interval [0, 2pi). sec x/2 = cos x/2

Sec(x/2) = 1/cos(x/2)

sec(x/2)=cos(x/2) > cos^2(x/2)=1 ---> cos(x/2) = -1 and cos(x/2) = 1

Cos(x/2)=1 --- > x/2 = 0, only.  x = 0;

cos(x/2)=-1 > x/2 = pi -> x = 2pi. But the statement says [0,2pi), so 2pi can not be chosen.

Only x = 0.

In fact, your equation is equivalent to sec(x)=cos(x), for x in [ 0, pi), so yes, only x = 0 .