Solving over an interval

[Hooty143]

sec(x)-sqrt(2)=0
interval is 0<=x<=2pi

steps I have done are

sec(x)=sqrt(2)

I know the answers are pi/4, 7pi/4 but I do not know how to get them!

 

[tkhunny]

Have you considered:

\(\displaystyle \cos(x)\;=\;\frac{\sqrt{2}}{2}\)

?

 

[Hooty143]

Thank you,

I know that sec(x)=1/cos(x) so would that mean 1/cos sqrt(2)/2?

 

[mmm4444bot]

steps I have done are

sec(x) = sqrt(2)

Your word "steps" must be a typographical error; I see only one step.

 

[mmm4444bot]

I know that sec(x) = 1/cos(x) so would that mean 1/cos sqrt(2)/2 ?

Your pronoun "that" makes no sense. You've asked if the equation in blue is the same statement as the expression in red. Of course, it is not, so you must be trying to ask something else.

Your first step is a good one. You added sqrt(2) to both sides.

sec(x) = sqrt(2)

Since you already know that sec(x) is the reciprocal of cos(x), make that substitution in your first result.

1/cos(x) = sqrt(2)

Take the reciprocal of both sides, to get cos(x) out of the denominator position. (In general, we do not solve for variables while they're in a denominator.)

cos(x) = 1/sqrt(2)

Rationalize the denominator.

cos(x) = sqrt(2)/2

Now, what angles between 0 and 2Pi have a cosine value of sqrt(2)/2 ?

You're done.

 

[Hooty143]

Thank you so much!! I get it now, I was definitly making that problem harder then it really was!