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engineering cal 2 - solve for x ---- arccos x = arcsec x
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I understand that y=arccos x and y=arcsec x but they are giving me the answer x =+/- 1
um, arcsec(x) = arccos(1/x) --> x = 1/x --> x^2 = 1 --> x = +-1
sec x = cos x ---> 1/cos x = cos x ---> 1 = cos x ^2 ----> +- 1 = cos x ??
is this new problem?
no no no I'm trying to make sense of your post. where did you get arccos (1/x)?
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but you are correct above....best way to see it is use unit circle, say x = cos(theta) --> theta = arccos(x) then sec(theta) = 1/x --> theta = arcsec(1/x)
yes sir. thank you very much
no problem
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