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For inverse trig functions.. Can someone describe how to find "arcsec2" or something similar? Thank you
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First let \[\large arcsec2 = x\]then that means \[\large 2 = \sec x\]Hopefully you know that sec x is the same as 1/cos x, so \[\large 2 = \frac{ 1 }{ \cos x }\]
Next you can solve for cos x, and then from there you need to make use of the unit circle to find x.
thank you, i'll come back to this in a sec
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