How do you evaluate #sec^-1(2)# without a calculator?

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Answer 1 · 2018-03-08T11:21:47

See below.

#secx =1/cosx#:

If:

#secx=2=>1/cosx=2#

#1/cosx=2=>cosx=1/2#

#x=arccos(cosx)=arccos(1/2)=pi/3 , (5pi)/3#

General solution:

#pi/3+2pin#

#(5pi)/3+2pin#

For integers #bbn#