How do you solve #5=sec^2x+3# for #0<=x<=2pi#?

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Answer 1 · 2016-09-12T02:35:35

#x=pi/4, (3pi)/4, (5pi)/4, (7pi)/4#

#5=sec^2x + 3#

#sec^2x = 2#

#secx =+-sqrt(2)#

#cosx =+-1/sqrt(2)#

Taking positive result:
# x= arccos(1/sqrt(2)) -> x=pi/4 or (7pi)/4# for #x in {0, 2pi)#

Taking negative result:
#x=arccos(-1/sqrt(2)) -> x= (3pi)/4 or (5pi)/4# for #x in {0, 2pi)#

Hence: #x=pi/4, (3pi)/4, (5pi)/4, (7pi)/4# for #x in {0, 2pi)#