<s>[INST] If $\tan 2\alpha \cdot \tan \alpha = 1$, then what is $\alpha$? Different methods give different answers. 
If $\tan 2\alpha\cdot\tan \alpha = 1$, then what is $\alpha$?

I tried two methods but got two different answers.
Method 1:
$$\begin{align}
\tan 2\alpha\cdot\tan \alpha = 1
&\implies
\frac{2\tan \alpha}{1 - \tan ^2 \alpha}\;\tan \alpha = 1 \tag{1a}\\[6pt]
&\implies 2\tan ^2\alpha = 1 - \tan ^2\alpha \tag{1b}\\[6pt]
&\implies \tan ^2 \alpha = \frac{1}{3} \tag{1c}\\[6pt]
&\implies \tan \alpha = \pm\frac{1}{\sqrt 3} \tag{1d}\\[6pt]
&\implies \tan \alpha = \tan\left(\pm\frac{\pi}{6}\right) \tag{1e}\\[6pt]
&\implies \alpha = n\pi \pm \frac{\pi}{6} \;\text{where}\; n \in \mathbb{Z} \tag{1f}
\end{align}$$
Method 2:
$$\begin{align}
\tan 2\alpha \cdot \tan \alpha = 1
&\implies \tan 2\alpha = \frac{1}{\tan \alpha} \tag{2a}\\[6pt]
&\implies \tan 2\alpha = \cot \alpha \tag{2b}\\[6pt]
&\implies \tan 2\alpha = \tan\left(\frac{\pi}{2} - \alpha\right) \tag{2c}\\[6pt]
&\implies 2\alpha = n\pi + \frac{\pi}{2} - \alpha \text{?} \tag{2d}\\[6pt]
&\implies \alpha = \frac{1}{3}\left(n\pi + \frac{\pi}{2}\right)\;\text{where}\; n \in \mathbb{Z} \tag{2e}
\end{align}$$
Which one is correct? Is there any mistake in the above solutions?
 [/INST] @Toky,
There is nothing wrong with your methods. Both are correct. You are getting two different representations for the same set of solutions. According to the second solution
$$
\dots -\frac{\pi}{6}, \frac{\pi}{6}, \frac{3\pi}{6}, \frac{5\pi}{6},\dots
$$
Your first solutions gives you half of these if you choose the $+$ sign, and gives you the other half if you choose the $-$ sign.
Hope this helps.
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