#sec^2x-1+tanx-sqrt(3)tanx=sqrt3# for#0 <= x <= 2pi#
Using the Pythagorean identity #tan^2x+1=sec^2x#, substitute for the #sec^2x# term.
#(tan^2x+1)-1+tanx-sqrt(3)tanx=sqrt3#
#tan^2x+tanx-sqrt(3)tanx=sqrt3color(white)(11aaa11)#Combine like terms
#tan^2x+tanx-sqrt3tanx-sqrt3=0color(white)(aaa)#Subtract #sqrt3# from both #color(white)(AaaaaaaaaaaaaaaaaaaaaaaAAAAAAAAAAA)#sides
#tan^2x +(1-sqrt3)tanx-sqrt3=0color(white)(aaaaa)#Group the 2 tan terms
#(tanx+1)(tanx-sqrt3)=0color(white)(aaaaaaaaa)#Factor
#tanx=-1color(white)(aaaaa)tanx=sqrt3color(white)(aaaaa)#Set factors = 0 and solve
#x=(3pi)/4#,#(7pi)/4color(white)(aaaaaa)x=(pi)/3#,#(4pi)/3color(white)(aaaa)#Use the unit circle*
*To find values for #tanx=-1# on the unit circle, #sinx# and #cosx# must be equal but with opposite signs. These conditions are found at #(3pi)/4# where #sinx=sqrt2/2#, #cosx=-sqrt2/2# and #tanx=sinx/cosx=-1#
and at #(7pi)/4# where #sinx=-sqrt2/2#, #cosx=sqrt2/2# and
#tanx=sinx/cosx=-1#
Similarly, to find values for #tanx=sqrt3# on the unit circle,
#tanx=sinx/cosx=sqrt3/2-:1/2=sqrt3# at #pi/3# and
#tanx=sinx/cosx=-sqrt3/2-:-1/2=sqrt3# at #(4pi)/3#
