Question 1128735


 If we restrict the domain of {{{t}}} to [{{{0}}},{{{2pi}}}) then 

{{{t=pi/3}}} or {{{5pi/3}}}


Explanation:

{{{2tan^2(t) = 3 sec(t) }}}

{{{2sin^2(t)/cos^2(t) = 3(1/cos(t) )}}}

{{{2sin^2(t) = 3 (cos^2(t)/cos(t) )}}}

{{{2sin^2(t) = 3 cos(t) }}}

{{{2(1-cos^2(t)) = 3 cos(t)}}}
 
{{{2 -2cos^2(t) = 3 cos(t)}}}
 
{{{2cos^2(t) +3 cos(t) -2=0}}} .......factor

{{{2cos^2(t) +4 cos(t)-cos(t) -2=0}}}

{{{(2cos^2(t) +4 cos(t))-(cos(t) +2)=0}}}

{{{2cos(t)(cos(t) +2)-(cos(t) +2)=0}}}

{{{(2cos(t)-1)(cos(t) +2)=0}}}

solutions:

case 1:

if {{{2cos(t)-1=0}}} ->{{{ cos(t)=1/2}}} => {{{t=pi/3}}} or {{{t=5pi/3}}} assuming [{{{0}}},{{{2pi}}})

case 2:

if {{{cos(t) +2)=0}}}->{{{cos(t) =-2}}}  BUT, {{{cos(t)}}} is element of [{{{-1}}},{{{1}}}] for {{{all}}}{{{ t}}}; therefore, this case is extraneous and we will disregard it