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jpjones

exact values for secx=-2/sqrt3

  • one year ago
  • one year ago

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

    • one year ago
  2. Mimi_x3
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    I think that you're able to do it now..

    • one year ago
  3. Mimi_x3
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    \[\frac{1}{cosx} = -\frac{2}{\sqrt{3}} \]

    • one year ago
  4. jpjones
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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

    • one year ago
  5. Mimi_x3
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    well cross multiply

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

    • one year ago
  7. jpjones
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    thanks ill keep trying, would i b using the fourth quadrant to draw this

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

    • one year ago
  9. jpjones
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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

    • one year ago
  10. jpjones
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    i think the angle has to be in one quadrant or the other

    • one year ago
  11. jpjones
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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

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

    • one year ago
  13. Mimi_x3
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    |dw:1344662088473:dw| \[ x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\]

    • one year ago
  14. Mimi_x3
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    wait something looks wrong

    • one year ago
  15. jpjones
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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

    • one year ago
  16. jpjones
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    that is but i gave you the answer to a different question originaly

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

    • one year ago
  18. jpjones
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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

    • one year ago
  19. Mimi_x3
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    what???

    • one year ago
  20. Mimi_x3
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    you just tricked me with a wrong answer?

    • one year ago
  21. jpjones
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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

    • one year ago
  22. jpjones
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    are you still there mimi x3

    • one year ago
  23. jpjones
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    come on mimi your my only hope here, where are you

    • one year ago
  24. nabsz.j
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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

    • one year ago
  25. nabsz.j
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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, \[x = \pi-(\pi/6)\] \[x= 5\pi/6\]

    • one year ago
  26. nabsz.j
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    aw man, how do you do fractions in this thing? i hate forward slashes as fractions -.-

    • one year ago
  27. Mimi_x3
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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{..}{..}

    • one year ago
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