How do you solve #2tan^2x - 3secx = 0#?

1 Answer
Alan P.
Jul 14, 2015

If we restrict the domain of #x# to #[0, 2pi)#
then #x=(pi)/3 or (5pi)/3#

Explanation:

#2tan^2(x) - 3sec(x) = 0#

#rarr 2(sin^2(x))/(cos^2(x)) - 3(1/cos(x)) = 0#

#color(white)("XXXX")#assuming #cos(x)!=0# (the equation would not be valid if it were)
#color(white)("XXXX")#Multiply by #cos^2(x)#
#2sin^2(x) = 3cos(x)#
#color(white)("XXXX")#Use Pythagorean relationship
#rarr 2(1-cos^2(x)) -3cos(x) = 0#
#color(white)("XXXX")#Simplify
#rarr 2cos^2(x)+3cos(x)-2=0#
#color(white)("XXXX")#Factor
#rarr (2cos(x)-1)(cos(x)+2) = 0#

#rArr 2cos(x)-1 = 0#
or
#color(white)("XXXX")cos(x)+2 = 0#

*Case 1
If #2cos(x)-1 = 0#
#color(white)("XXXX")rarr cos(x) = 1/2#
#color(white)("XXXX")rarr x = (pi)/3 or (5pi)/3#
#color(white)("XXXX")color(white)("XXXX")#assuming #x in [0,2pi)#

Case 2
If #cos(x)+2 = 0#
#color(white)("XXXX")rarr cos(x) = -2#
BUT
#color(white)("XXXX")cos(x) in [-1,+1] " for all " x#
Therefore this case is extraneous.

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