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jpjones Group Title

exact values for secx=-2/sqrt3

  • 2 years ago
  • 2 years ago

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  1. Mimi_x3 Group Title
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    \[secx = -\frac{2}{\sqrt{3}} \] \[secx = \frac{1}{cosx} \]

    • 2 years ago
  2. Mimi_x3 Group Title
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    I think that you're able to do it now..

    • 2 years ago
  3. Mimi_x3 Group Title
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    \[\frac{1}{cosx} = -\frac{2}{\sqrt{3}} \]

    • 2 years ago
  4. jpjones Group Title
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    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

    • 2 years ago
  5. Mimi_x3 Group Title
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    well cross multiply

    • 2 years ago
  6. Mimi_x3 Group Title
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    \[\frac{1}{cosx} = \frac{-2}{\sqrt{3}} => \sqrt{3} = -2cosx\] you should be able to do it now :)

    • 2 years ago
  7. jpjones Group Title
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    thanks ill keep trying, would i b using the fourth quadrant to draw this

    • 2 years ago
  8. Mimi_x3 Group Title
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    |dw:1344660849255:dw| since cosx is a negative; so you will use the quadrants that i marks

    • 2 years ago
  9. jpjones Group Title
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    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

    • 2 years ago
  10. jpjones Group Title
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    i think the angle has to be in one quadrant or the other

    • 2 years ago
  11. jpjones Group Title
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    oops i gave you the wrong answer a moment ago, the correct answer is pi/3 and -2pi/3, maybe that will help

    • 2 years ago
  12. Mimi_x3 Group Title
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    hmm; i dont know how to explain sorry \[ cosx = -\frac{\sqrt{3}}{2} => x = \frac{\pi}{6} + \pi => \frac{7\pi}{6} \]

    • 2 years ago
  13. Mimi_x3 Group Title
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    |dw:1344662088473:dw| \[ x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\]

    • 2 years ago
  14. Mimi_x3 Group Title
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    wait something looks wrong

    • 2 years ago
  15. jpjones Group Title
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    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

    • 2 years ago
  16. jpjones Group Title
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    that is but i gave you the answer to a different question originaly

    • 2 years ago
  17. Mimi_x3 Group Title
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    i doubled checked; i dont think i made any mistake...wait you have me the answers from a different question?

    • 2 years ago
  18. jpjones Group Title
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    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

    • 2 years ago
  19. Mimi_x3 Group Title
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    what???

    • 2 years ago
  20. Mimi_x3 Group Title
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    you just tricked me with a wrong answer?

    • 2 years ago
  21. jpjones Group Title
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    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

    • 2 years ago
  22. jpjones Group Title
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    are you still there mimi x3

    • 2 years ago
  23. jpjones Group Title
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    come on mimi your my only hope here, where are you

    • 2 years ago
  24. nabsz.j Group Title
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    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

    • 2 years ago
  25. nabsz.j Group Title
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    \[\sec (x) = -2/\sqrt{3} \] \[1/\cos(x) = -2/\sqrt{3}\] 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, \[x = \pi-(\pi/6)\] \[x= 5\pi/6\]

    • 2 years ago
  26. nabsz.j Group Title
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    aw man, how do you do fractions in this thing? i hate forward slashes as fractions -.-

    • 2 years ago
  27. Mimi_x3 Group Title
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    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{..}{..}

    • 2 years ago
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