How do you solve #4secx+8=0#?

2 Answers
Deepak G.
Aug 9, 2016

#x=120^@=(2pi)/3#

Explanation:

#4secx+8=0#
or
#4secx=-8#
or
#secx=-8/4#
or
#secx=-2#
or
#1/cosx=-2#
or
#cosx=-1/2#
or
#cosx=cos120^@#
or
#x=120^@=(2pi)/3#

Nghi N.
Aug 9, 2016

#x = (2pi)/3 + 2kpi#
#x = (4pi)/3 + 2kpi#

Explanation:

#sex x = 1/(cos x) = -8/4#
#cos x = -4/8 = - 1/2#
Trig table of special arcs, and trig unit circle -->
#cos x = - 1/2# --> arc #x = +- (2pi)/3#
The arc #-(2pi)/3# is the same as the arc #(4pi)/3# --> co-terminal arcs.
General answers:
#x = (2pi)/3 + 2kpi#
#x = (4pi)/3 + 2kpi#

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