How do you solve #4tanx=2sec^2x# for #0<=x<=2pi#?

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Answer 1 · 2016-10-20T02:22:11

#x=pi/4# and #x=(5pi)/4#

#(4tanx)/2=(2sec^2x)/2#

#2tanx=sec^2x#

Use the Pythagorean identity #sec^2x=tan^2x+1#

#2tanx=tan^2x+1#

#0=tan^2x-2tanx+1color(white)(aaa)#Subtract #2tanx# from both sides

#0=(tanx-1)(tanx-1)color(white)(aaa)#Factor

#tanx-1=0color(white)(aaa)#Set the factors equal to zero

#tanx=1color(white)(aaaaa)#Add #1# to both sides

For #0<=x<=2pi#, #tanx=1# where #sinx=cosx#
(because #tanx=sinx/cosx#).

Use the unit circle to find angles where
#sinx=cosx#.

#tanx=1# at #x=pi/4# and #x=(5pi)/4#