How do you solve #2-10secx=4-9secx# for #0<=x<=2pi#?

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Answer 1 · 2016-11-13T20:41:10

#x=(2pi)/3, x=(4pi)/3#

#2-10secx=4-9secx#

Isolate #secx#:
#2-secx=4#
#-secx=2#

#secx=-2#

To solve for x, I find it easier when I'm working in terms of sin, cos, or tan rather than sec, so I'll convert it:

#1/cosx=-2#

#-2cosx=1#
#cosx=-1/2#

*Remember our domain is restricted to #0<=x<=2pi#
#x=(2pi)/3, x=(4pi)/3#