The general solution of the equation $\tan \,2\theta\cdot \tan\,\theta=1$ for $n\in z$ is

$\begin{aligned} & \tan 2 \theta \cdot \tan \theta=1 \\ \Rightarrow & \left(\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right) \cdot \tan \theta=1 \\ \Rightarrow & \frac{\left(2 \tan ^{2} \theta\right)}{\left(1-\tan ^{2} \theta\right)}=1 \\ \Rightarrow & 2 \tan ^{2} \theta=1-\tan ^{2} \theta \\ \Rightarrow & 3 \tan ^{2} \theta=1 \\ \Rightarrow & \tan ^{2} \theta=\frac{1}{3} \\ \Rightarrow & \tan \theta= \pm \sqrt{\frac{1}{3}} \\ \therefore & \theta=n \pi \pm \frac{\pi}{6}\end{aligned}$