Question

4tantheta/12-tan^2=1

Answer 1

$4tan \theta = 12 - tan^{2} \theta$
$tan^{2} \theta + 4tan \theta - 12 = 0$

Let $x = tan \theta$
$x^{2} + 4x - 12 = 0$
$(x + 6)(x - 2) = 0$
$x = -6, x = 2$

$tan \theta = -6, tan \theta = 2$

Since they both work, then we can write the solutions in general form:
$\theta = n \pi + tan^{-1}(2)$, $n \in Z$
$\theta = n \pi - tan^{-1}(6)$, $n \in Z$

Now, these solutions will always satisfy the equation and substituting a whole number in place of n will produce a solution that satisfies the given equation.