Question 1178808

given: {{{tan(x) +sec(x) =sqrt(3)}}}  and {{{0< x < 2pi}}}


We use the identity: {{{sin^2(x) +cos^2( x )= 1}}}  to find {{{cos(x)}}}


{{{cos^2( x )= 1-sin^2(x) }}}..........factor


{{{cos^2( x )=( 1-sin(x))( 1+sin(x))}}}


{{{cos^2( x )/( 1-sin(x))=( 1+sin(x))}}}.....1)



{{{tan(x)=sin(x)/cos(x)}}} and {{{sec(x)=1/cos(x)}}}



{{{sin(x)/cos(x) +1/cos(x) =sqrt(3)}}}


{{{(sin(x)+1)/cos(x) =sqrt(3)}}}.......substitute {{{( 1+sin(x))}}} from 1)


{{{(cos^2( x )/( 1-sin(x)))/cos(x) =sqrt(3)}}}


{{{cos(x)/(1 - sin(x))=sqrt(3)}}}


use the identity: {{{sin^2(x) +cos^2( x )= 1}}}  to find {{{sin(x)}}}


{{{sin(x)=sqrt(1-cos^2( x ))}}}


{{{cos(x)/(1 - sqrt(1-cos^2( x )))=sqrt(3)}}}


{{{cos(x)=(1 - sqrt(1-cos^2( x )))*sqrt(3)}}}


{{{cos(x)/sqrt(3)=1 - sqrt(1-cos^2( x ))}}}


{{{cos(x)/sqrt(3)-1= - sqrt(1-cos^2( x ))}}}.....square both sides


{{{(cos(x)/sqrt(3)-1)^2= (- sqrt(1-cos^2( x )))^2}}}


{{{(cos(x)-sqrt(3))/(sqrt(3))^2= 1-cos^2( x )}}}


{{{(cos(x)-sqrt(3))^2/3= 1-cos^2( x )}}}


{{{cos^2(x)-2cos(x)*sqrt(3)+3= 3-3cos^2( x )}}}


{{{cos^2(x)+3cos^2( x )-2cos(x)*sqrt(3)+3- 3=0}}}


{{{4cos^2( x )-2cos(x)*sqrt(3)=0}}}


{{{2cos(x)(2cos( x )-sqrt(3))=0}}}


solutions:


if {{{2cos(x)=0}}}=> {{{cos(x)=0}}}

or

if {{{2cos( x )-sqrt(3)=0}}}=>{{{2cos( x )=sqrt(3)}}}=>{{{cos( x )=sqrt(3)/2}}}



Since {{{tan (x)}}} and {{{cos (x)}}} are positive, then {{{x}}} lies in the first quadrant, and in {{{sin(x)/cos(x) +1/cos(x) =sqrt(3)}}}


{{{cos(x)=0}}} is denominator, so disregard this solution


{{{cos (x )= sqrt(3)/2}}} 


{{{x= cos^-1(sqrt(3)/2 )}}}


{{{x= pi/6}}} (result in radians)-> your solution


convert to degrees


{{{ x = 30}}}° (degrees)