Question 622504
{{{4(tan(u))^2-1=(tan(u))^2}}}
Let me change the name of {{{tan(u)}}} for a little while, to save ink and confusion.
{{{tan(u)=x^2}}}
Now I can write the confusing equation as
{{{4x^2-1=x^2}}} --> {{{3x^2-1=0}}} --> {{{3x^2=1}}} --> {{{x^2=1/3}}}
Back into trigonometry,
{{{(tan(u))^2=1/3}}}
You should know an angle that is a solution.
(You should sine and cosine of 0, 30,45,60, and 90 degrees, or at least have a chart with their exact values handy).
{{{30^o}}} or {{{pi/6}}} is a solution.
{{{sin(30^o)=1/2}}}, {{{cos(30^o)=sqrt(3)/2}}}, and {{{tan(30^o)=sin(30^o)/cos(30^o)=1/sqrt(3)}}}
(although teachers prefer the equivalent expression {{{tan(30^o)=sqrt(3)/3}}} because they do not like square roots in denominators).
You also should know that the supplementary angle, {{{150^o}}} or {{{5pi/6}}} has the opposite tangent:
{{{tan(150^o)=-1/sqrt(3)}}}
So those two angles are solutions.
There are many more solutions, because the tangent function has a period of {{{180^o}}} or {{{pi}}}, so to include all solutions we need to add {{{n*pi}}} with {{{n}}} defined as an integer to the solutions above.
(I am not going to continue with the degrees, because I'm sure your teacher likes pies better than degrees).
In radians, and putting all solutions in one formula, my solution set is
{{{u=n*pi +- pi/6}}} for all n integers