How do you solve #sec 4x = 2# in the interval 0 to 2pi?

1 Answer
Alan P.
Feb 12, 2016

#x in {pi/12, (5pi)/12, (7pi)/12,(11pi)/12,(13pi)/12,(17pi)/12,(19pi)/12,(23pi)/12}#

Explanation:

If #sec(4x) = 2#
then
#color(white)("XXX")4x=2pi*k+-pi/3# for #kinZZ#
#color(white)("XXXXXXXXXX")#this is based on the standard angles:
#color(white)("XXXXXXXXXX")cos(+-pi/3)=1/2#

#x=(pik)/2+-pi/12 = pi(6k+-1)/12#
#color(white)("XXX")#for all values of #k# which evaluate as #xin[0,2pi]#

#k=0 rarr x=pi/12# (other values give #x < 0#)

#k=1 rarr x= (5pi)/12 or (7pi)/12#

#k=2 rarr x=(11pi)/12 or (13pi)/12#

#k=3 rarr x=(17pi)/12 or (19pi)/12#

#k=4 rarr x=(23pi)/12# (other values give #x > 0#)

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