Question

How do you find `sec x= sqrt2`?

Answer 1

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`.

Answer 2

`x=pi/4,(7*pi)/4`

Explanation:

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`