How do you solve #3sec^2x - 4=0#?

1 Answer
Lucy
Jul 25, 2018

#x=pi/6,(5pi)/6,(7pi)/6,(11pi)/6# in the domain #0 <=x <=2pi#

Explanation:

#3sec^2x=-4=0#
#sec^2x=4/3#
#1/cos^2x=4/3#
#cos^2x=3/4#
#cosx=+-sqrt3/2#
#x=pi/6,pi-pi/6,pi+pi/6,2pi-pi/6#
#x=pi/6,(5pi)/6,(7pi)/6,(11pi)/6# in the domain #0 <=x <=2pi#

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