- anonymous

exact values for secx=-2/sqrt3

- jamiebookeater

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- Mimi_x3

\[secx = -\frac{2}{\sqrt{3}} \]
\[secx = \frac{1}{cosx} \]

- Mimi_x3

I think that you're able to do it now..

- Mimi_x3

\[\frac{1}{cosx} = -\frac{2}{\sqrt{3}} \]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

no you'll have to explain a bit better than that, i can tell you the answer is 5pi/6 and -5pi/6 but im not sure how to get there

- Mimi_x3

well cross multiply

- Mimi_x3

\[\frac{1}{cosx} = \frac{-2}{\sqrt{3}} => \sqrt{3} = -2cosx\]
you should be able to do it now :)

- anonymous

thanks ill keep trying, would i b using the fourth quadrant to draw this

- Mimi_x3

|dw:1344660849255:dw|
since cosx is a negative; so you will use the quadrants that i marks

- anonymous

that getting clearer for me, now if you could just name the angles with the answers i gave before i might be able to complete the puzzle

- anonymous

i think the angle has to be in one quadrant or the other

- anonymous

oops i gave you the wrong answer a moment ago, the correct answer is pi/3 and -2pi/3, maybe that will help

- Mimi_x3

hmm; i dont know how to explain sorry
\[ cosx = -\frac{\sqrt{3}}{2} => x = \frac{\pi}{6} + \pi => \frac{7\pi}{6} \]

- Mimi_x3

|dw:1344662088473:dw|
\[ x = \pi
- \frac{\pi}{6}
= \frac{5\pi}{6}\]

- Mimi_x3

wait something looks wrong

- anonymous

yes have a look back a couple of replies the answer is pi/3 and -2pi/3, once again i am sorry bi gave you the answer to a different question originally

- anonymous

that is but i gave you the answer to a different question originaly

- Mimi_x3

i doubled checked; i dont think i made any mistake...wait you have me the answers from a different question?

- anonymous

i gotta tell you that is an amazing effort to come up with a correct answer to a question that i gave the incorrect answer to

- Mimi_x3

what???

- Mimi_x3

you just tricked me with a wrong answer?

- anonymous

i promise it was an accident im working on 4 equations at a time here, cramming for an exam in the following week, but now that i gave you the correct answer pi/3 and -2pi/3 it should be easy for you to show me, once again im sorry for that

- anonymous

are you still there mimi x3

- anonymous

come on mimi your my only hope here, where are you

- anonymous

can i know the question and the answer please because I don't wanna stuff up because of the confusion up above lol i might be able to help

- anonymous

\[\sec (x) = -2/\sqrt{3} \]
\[1/\cos(x) = -2/\sqrt{3}\]
Flip both sides of the equation to get cos(x) on top:
\[\cos(x) = -\sqrt{3}/2\]
Take arccos of both sides, you get \[x=\cos^{-1} (-\sqrt{3}/2)\]
\[x=-\pi/6\]
Then you know that the range and domain of arccos is only between 0 and pi so that's all the angles you are looking for. And we also know cos is negative so it must be in 2nd or 3rd quadrant.
So that leaves us with the 2nd quadrant and the answer, \[x = \pi-(\pi/6)\]
\[x= 5\pi/6\]

- anonymous

aw man, how do you do fractions in this thing? i hate forward slashes as fractions -.-

- Mimi_x3

ohh no no. the answer the i typed above is correct. i dont think that its allowed to be (-pi/6)
and for fractions \frac{..}{..}

Looking for something else?

Not the answer you are looking for? Search for more explanations.