Solve $3\tan^2\theta - 1 =0$. I have the question:
Solve for $\theta$ given $$3\tan^2θ-1=0$$
for $θ\in(-\pi ,\pi)$.
Anyone know how to do this? Any help would be greatly appreciated.
 A: $$\tan^2\theta=1/3\iff\cos2\theta=\frac{1-1/3}{1+1/3}=1/2=\cos\frac\pi3$$
$$\implies2\theta=2m\pi\pm\frac\pi3=\frac\pi3(6m\pm1)\iff\theta=\frac\pi6(6m\pm1)$$ where $m$ is any integer
We need $-\pi<\dfrac\pi6(6m\pm1)<\pi\iff-6<6m\pm1<6$
Check for '+','-' one by one.
A: $$3\tan^2\theta - 1 =0$$
$$\tan^2\theta=\frac 13$$
$$\tan\theta=\pm\frac{\sqrt 3}3$$
For angles in the first quadrant, only $\frac\pi 6$ has a tangent of $\frac{\sqrt 3}3$, so that is our reference angle for our solutions. We then find the corresponding angles in the other quadrants that have a tangent with the same absolute value. Since we want just the same absolute value of tangent, all of those angles will solve our equation.
In quadrant 1, our angle is just the reference angle, $\frac\pi 6$.
In quadrant 2, we want $\pi-\frac\pi 6=\frac{5\pi}6$.
In quadrant 3, we want $\frac\pi 6-\pi=-\frac{5\pi}6$.
In quadrant 4, we want $-\frac{5\pi}6$.
So our solution set is $\left\{-\frac{5\pi}6,-\frac{\pi}6,\frac{\pi}6,\frac{5\pi}6\right\}$. We can confirm this with a graph.

The $x$-axis tick marks are at every $\frac{\pi}6$, so this confirms our answer.
