How do you solve ##sec^2x-2=0##?
February 20th, 2023
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##