How do you find #sec x= sqrt2#?

Recall the definition of #secx# in terms of ratios of sides of a right-angled triangle: #secx=(hyp)/(adj)# - hypotenuse divided by adjacent sides.

Consider a triangle with other angles both #45^o#. The two sides adjoining the right-angle are of equal length, call it #a#. Pythagoras' Theorem tells us that the hypotenuse is of length #sqrt(a^2+a^2)=asqrt2#. So #sec45^o=(asqrt2)/a=sqrt2#.

If this doesn't seem intuitive due to the use of the less usual #sec# trig function, consider the definition of #secx# as #secx=1/cosx#. #secx=sqrt2rArrcosx=1/sqrt(2)#, one of the first cosine values that one learns: #cos45^o=1/sqrt2#.

This can be more simply interpreted if you change #secx# to #1/cosx#.

Then by manipulating the equation you can turn:
#1/cosx=sqrt(2)#

Into:
#cosx=1/sqrt(2)#

Since you cannot have radicals in the denominator, you multiply #1/sqrt(2)# by #sqrt(2)/sqrt(2)#

Giving you: #sqrt(2)/2#

From here, you go back to the unit circle. Where does the x value (due to cosine) equal #sqrt(2)/2#?

When #theta# (in this case x) is equal to #pi/4#

Since #secx=sqrt(2)# is positive and you are looking for an x value, you must remember that quadrants 1 and 4 have positive x values.

knowing that your answer is #theta#(again, in this case x)#=pi/4#, assuming that this is being taken from #[0,2pi]#, makes your two answers #pi/4# and #(7*pi)/4#