How do you solve # (1/(cos^2 x)) + 1 - 3 tan^2 x = 0#?

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Answer 1 · 2015-05-31T00:06:41

#(1/cos^2x)+1−3tan^2x=0#

#sec^2(x)+1-3tan^2x=0#

Since #color(blue)(tan^2x+1=sec^2x)#
then #color(blue)(1= sec^2x-tan^2x)#

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

#2sec^2x-4tan^2x = 0#

Factor and divide out #2#:

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

#sec^2x-2tan^2x=0#

#1/cos^2x = (2sin^2x)/(cos^2x)#

#1=2sin^2x#

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

Without restriction:
#x=pi/4 +kpi#

From #0 to 2pi#
#x = pi/4, (3pi)/4, (5pi)/4, (7pi)/4#

You can see it on the graph:

graph{(secx)^2+1-3(tanx)^2 [-10.125, 9.875, -2.28, 7.72]}