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exact values for secx=-2/sqrt3

Mathematics
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\[secx = -\frac{2}{\sqrt{3}} \] \[secx = \frac{1}{cosx} \]
I think that you're able to do it now..
\[\frac{1}{cosx} = -\frac{2}{\sqrt{3}} \]

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Other answers:

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
well cross multiply
\[\frac{1}{cosx} = \frac{-2}{\sqrt{3}} => \sqrt{3} = -2cosx\] you should be able to do it now :)
thanks ill keep trying, would i b using the fourth quadrant to draw this
|dw:1344660849255:dw| since cosx is a negative; so you will use the quadrants that i marks
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
i think the angle has to be in one quadrant or the other
oops i gave you the wrong answer a moment ago, the correct answer is pi/3 and -2pi/3, maybe that will help
hmm; i dont know how to explain sorry \[ cosx = -\frac{\sqrt{3}}{2} => x = \frac{\pi}{6} + \pi => \frac{7\pi}{6} \]
|dw:1344662088473:dw| \[ x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\]
wait something looks wrong
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
that is but i gave you the answer to a different question originaly
i doubled checked; i dont think i made any mistake...wait you have me the answers from a different question?
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
what???
you just tricked me with a wrong answer?
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
are you still there mimi x3
come on mimi your my only hope here, where are you
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
\[\sec (x) = -2/\sqrt{3} \] \[1/\cos(x) = -2/\sqrt{3}\] 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, \[x = \pi-(\pi/6)\] \[x= 5\pi/6\]
aw man, how do you do fractions in this thing? i hate forward slashes as fractions -.-
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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