Notice, before proceeding, that we must have #cosx != 0# due to the presence of #2/cosx# in the equation.
#cosx-1 = 2/cosx#
#=> cosx(cosx-1) = cosx(2/cosx)#
#=> cos^2x - cosx = 2#
#=> cos^2x - cosx - 2 = 0#
#=> (cosx+1)(cosx-2) = 0#
#=> cosx = -1# or #cosx = 2#
As #|cosx|<=1# for #x in RR#, #cosx=2# has no real solutions. Thus we are left with #cosx=-1# as the only option.
#cosx = -1#
#=> x = pi+2pin, n in ZZ#.
As we have the restriction #x in (0, 2pi)#, the only choice for #n# which works is #n=0#. Thus we get our solution:
#x = pi#
Checking, we find that it satisfies the given equation and does not violate the restriction of #cosx != 0# we set in the beginning, as desired.
