These two methods of solving the $\tan(2\theta + \frac{\pi}{4})$ ，which one is correct? I just prefer the Method 2 because of the period, but I don’t know how could Method 1 is incorrect if it comes out with the same answer as Method 2 does.
What should I write down if I get an answer which does not in the domain that the question require? $\theta = \frac{11\pi}{24}$ (unqualified)?
 A: The general idea may help you to choose.
$$\tan\left(2\theta+\frac{\pi}{4}\right)=\tan\left(\frac{\pi}{6}\right)\\
\implies 2\theta+\frac{\pi}{4}=k\pi+\frac{\pi}{6}\\\implies
\theta+\frac{\pi}{8}=k\frac{\pi}{2}+\frac{\pi}{12}\\\implies
\theta=k\frac{\pi}{2}-\frac{\pi}{24}\\\implies0\leq \theta \leq \pi\\\implies
0\leq k\frac{\pi}{2}-\frac{\pi}{24} \leq \pi\\\implies k=1,2.$$
A: Notice that $\tan(2\theta+\pi/4)$ is a periodic function, with period equal to $\pi/2$. So if you have a solution outside the domain, you can add (or subtract) multiples of $\pi/2$ to get it in the desired range.
So if $\theta=-\pi/24$ is a solution outside the range, add $\pi/2$ or $2\pi/2=\pi$ to get it in the $[0,\pi]$ range.
A: Neither.  Before restricting to the interval, you should have an infinite family of solutions, all separated by half(!)(footnote) the period of tangent.
\begin{align*}
\tan \left( 2\theta + \frac{\pi}{4} \right) &= \frac{1}{\sqrt{3}}  \text{,} \\
2\theta + \frac{\pi}{4}  &=  \arctan\left( \frac{1}{\sqrt{3}} \right) + \pi k, k \in \Bbb{Z} \\
&=  \frac{\pi}{6} + \pi k, k \in \Bbb{Z}  \text{,}  \\
2\theta  &=  \frac{\pi}{6} - \frac{\pi}{4} + \pi k, k \in \Bbb{Z} \\
  &=  \frac{-\pi}{12}  + \pi k, k \in \Bbb{Z}  \text{, and}  \\
\theta  &=  \frac{1}{2} \left( \frac{-\pi}{12}  + \pi k \right), k \in \Bbb{Z} \\
  &=  \frac{-\pi}{24}  + \frac{\pi}{2} k, k \in \Bbb{Z} \text{.}
\end{align*}
So, as $k$ ranges through the integers, every angle $-\pi/24+ (\pi/2)k$ yields a value for $\theta$ solving the equation.  We wish to restrict to angles in the interval $[0,\pi]$, so there is no need to consider negative $k$s further.  We make a table and for this, it is convenient to rewrite the "$\pi/2$" with denominator $24$ to match the angle: $-\pi/24 + (12\pi/24)k$.
\begin{align*}
k &= 0  &  \theta &= -\pi/24  \\
k &= 1  &  \theta &= 11\pi/24  \\
k &= 2  &  \theta &= 23\pi/24  \\
k &= 3  &  \theta &= 35\pi/24  \\
\end{align*}
Clearly, increasing $k$ further gives no more solutions in the interval.  Therefore, the solutions in the interval are $\theta = 11\pi/24$ and $\theta = 23\pi/24$.
(footnote)  Why half?  Because $\theta$ appears multiplied by $2$.  Note that $\tan(2\theta)$ has period $\pi/4$.  If you just treat the period as some afterthought to your computation, you are very likely to make the error made in your work and fail to find all the solutions.  If you carry the period through from the instant you apply inverse functions (and remember that multiplying or dividing both sides of the equation applies to both terms on the right), you don't have to make any special adjustments in your handling of the period based on what $\theta$'s coefficient is.
A: Due to the periodicity of the function $\tan$ (just examine its graph), in general for $n\in\mathbb Z,$ $$\tan\alpha=\tan\beta\iff\alpha=n\pi+\beta.$$
So for $n\in\mathbb Z,$
\begin{equation}
\begin{aligned}
\tan\left(2\theta+\frac{\pi}4\right) &=\frac1{\sqrt3}\\
\iff\tan\left(2\theta+\frac{\pi}4\right) &=\tan\left(\tan^{-1}\left(\frac1{\sqrt3}\right)\right)\\
\iff\tan\left(2\theta+\frac{\pi}4\right) &=\tan\left(\frac{\pi}6\right)\\
\iff2\theta+\frac{\pi}4 &= n\pi+\frac{\pi}6\\
\iff\theta &= \frac{\pi}{24}\left(12n-1\right).
\end{aligned}
\end{equation}
If $\theta\in[0,\pi],\;$ then $n=1$ or $2\,$ and $\,\theta=\frac{11\pi}{24}$ or $\frac{23\pi}{24}.$
