SOLUTION: please help me solve: 4tan^2u-1= tan^2u

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Question 622504: please help me solve:
4tan^2u-1= tan^2u

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
4%28tan%28u%29%29%5E2-1=%28tan%28u%29%29%5E2
Let me change the name of tan%28u%29 for a little while, to save ink and confusion.
tan%28u%29=x%5E2
Now I can write the confusing equation as
4x%5E2-1=x%5E2 --> 3x%5E2-1=0 --> 3x%5E2=1 --> x%5E2=1%2F3
Back into trigonometry,
%28tan%28u%29%29%5E2=1%2F3
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%5Eo or pi%2F6 is a solution.
sin%2830%5Eo%29=1%2F2, cos%2830%5Eo%29=sqrt%283%29%2F2, and tan%2830%5Eo%29=sin%2830%5Eo%29%2Fcos%2830%5Eo%29=1%2Fsqrt%283%29
(although teachers prefer the equivalent expression tan%2830%5Eo%29=sqrt%283%29%2F3 because they do not like square roots in denominators).
You also should know that the supplementary angle, 150%5Eo or 5pi%2F6 has the opposite tangent:
tan%28150%5Eo%29=-1%2Fsqrt%283%29
So those two angles are solutions.
There are many more solutions, because the tangent function has a period of 180%5Eo or pi, so to include all solutions we need to add n%2Api 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%2Api+%2B-+pi%2F6 for all n integers