## jpjones 3 years ago exact values for secx=-2/sqrt3

1. Mimi_x3

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

2. Mimi_x3

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

3. Mimi_x3

$\frac{1}{cosx} = -\frac{2}{\sqrt{3}}$

4. jpjones

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

5. Mimi_x3

well cross multiply

6. Mimi_x3

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

7. jpjones

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

8. Mimi_x3

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

9. jpjones

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

10. jpjones

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

11. jpjones

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

12. Mimi_x3

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

13. Mimi_x3

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

14. Mimi_x3

wait something looks wrong

15. jpjones

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

16. jpjones

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

17. Mimi_x3

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

18. jpjones

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

19. Mimi_x3

what???

20. Mimi_x3

you just tricked me with a wrong answer?

21. jpjones

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

22. jpjones

are you still there mimi x3

23. jpjones

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

24. nabsz.j

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

25. nabsz.j

$\sec (x) = -2/\sqrt{3}$ $1/\cos(x) = -2/\sqrt{3}$ 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, $x = \pi-(\pi/6)$ $x= 5\pi/6$

26. nabsz.j

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

27. 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{..}{..}

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