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

Question

↳Redirected from "How do you evaluate #sin^-1 (-sqrt3/2)#?"

4554 views around the world

You can reuse this answer
Creative Commons License

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#