Answer
$u=\frac{\pi}{6}+n\pi$ and $u=\frac{5\pi}{6}+n\pi$, where $n$ is an integer.
Work Step by Step
$4~tan^2u-1=tan^2u$
$3~tan^2u=1$
$tan^2u=\frac{1}{3}$
$tan~u=±\frac{1}{\sqrt 3}=±\frac{\sqrt 3}{3}$
The period of $tan~u$ is $\pi$. The solutions in the interval $[0,\pi)$ are:
$u=\frac{\pi}{6}$ and $u=\frac{5\pi}{6}$
Now, add multiples of $\pi$ to the solutions:
$u=\frac{\pi}{6}+n\pi$ and $u=\frac{5\pi}{6}+n\pi$, where $n$ is an integer.
