How do you solve #secx/cosx - 1/2secx= 0#?

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Answer 1 · 2015-08-20T12:48:37

There are no real solutions

# sec A = 1/cos A #

# sec x/cos x - 1/2sec x = 0 #
# sec x(sec x - 1/2) = 0 #
# sec x = 0 # or # sec x = 1/2#

The range of #|sec x| >= 1#, so there are no real solutions

graph{sec x [-10, 10, -5, 5]}

Answer 2 · 2015-08-26T19:35:01

Maple software produces this answer

#x=arccos(2)#

#x# is approximately #1.317i#

NOTE: This is not a real solution as the other poster dani83 states. If you are restricting yourself to the real numbers then dani83 is correct.

Rewrite
NOTE: #sec(x)=1/cos(x)#

#1/cosx(1/cosx)-1/2sec(x)=0#

This is just

#sec^2x-1/2secx=0#

Factor

#secx(secx-1/2)=0#

#secx# cannot be zero so

#secx-1/2=0#

#secx=1/2#

Take reciprocal

#cosx=2#

#arccos(cosx)=arccos(2)#

#x=arccos(2)#

This is not a real solution as the other poster says