How do you solve ##sec^2x-2=0##?

The general solution is ##npi+-pi/3##

As ##sec^2x-2=0##, we have

##sec^2x-(sqrt2)^2=0## or

##(secx+sqrt2)(secx-sqrt2)=0## and

##secx=+-sqrt2=+-sec(pi/3)##

Hence in the range ##(-pi,+pi)## the values are ##{+-pi/3,+-(2pi)/3}##, which can also be written as ##{+-pi/3,+-(pi-pi/3)}##

Hence, the general solution is ##npi+-pi/3##