How do you find all solutions of the equations #sec^2x-1=0# in the interval #[0,2pi)#?
1 Answer
Jun 17, 2017
Explanation:
#"using "#
#• secx=1/cosx#
#rArr1/cos^2x-1=0#
#rArrcos^2x=1#
#rArrcosx=+-1#
#cosx=1rArrx=cos^-1(1)=0#
#cosx=-1rArrx=cos^-1(-1)=pi#
#rArrx=0,pito[0,2pi)#
