Different ways gives different results - solving $\tan 2a = \sqrt 3 $

Question

Different ways gives different results - solving $\tan 2a = \sqrt 3$

Case 1).

$$ \tan 2a = \sqrt 3 =\tan(\frac{\pi}{3}) $$

$$2a = n\pi + \frac{\pi}{3} $$

$$a = \frac{n\pi}{2} + \frac{\pi}{6},\qquad n \in\mathbb{Z} $$

Case 2)

$$ \dfrac{2\tan a}{1-\tan^2 a} = \sqrt 3$$ Solving above equation gives $ \tan a = \dfrac{1}{\sqrt 3}$ or $-\sqrt 3 $

Thus either

$$ a = n\pi+ \frac{\pi}{6},\qquad n \in \mathbb{Z} $$

Or $$ a = m\pi - \frac{\pi}{3},\qquad m\in\mathbb{Z} $$

Or either I did something wrong or both these results are same.

If it's the later, I tried a lot to convert one form to other by adding the case 2 results. But that doesn't work.

Answer 1

$n\pi/2+\pi/6=m\pi+\pi/6\iff n=2m$

$n\pi/2+\pi/6=m\pi-\pi/3\iff3n+1=6m-2\iff n=2m-1$

Answer 2

They are the same. Your first solution is at all whole multiples of π plus either 1/6 π or 4/6 π (that is, a whole multiple of π, plus a half π, plus 1/6 π.)

Your second solution is at all whole multiples of π plus 1/6 π or plus 2/3 π (since $m\pi - \frac 1 3 \pi = (m-1)\pi + \frac 2 3 \pi$.)

Answer 3

If you are in doubt, enumerate the solutions.

Case 1:

$$\cdots-\frac{2\pi}6,\frac\pi6,\frac{4\pi}6,\frac{7\pi}6,\frac{10\pi}6,\frac{13\pi}6,\cdots$$

Case 2:

$$\cdots\frac\pi6,\frac{7\pi}6,\frac{13\pi}6,\cdots$$

and

$$\cdots-\frac{2\pi}6,\frac{4\pi}6,\frac{10\pi}6,\cdots$$