general solution beta $(\sin(\alpha+\beta)+\cos(\alpha +2\beta)\sin(\beta))^2 = 4\cos(\alpha)\sin(\beta)\sin(\alpha+ \beta);\tan(\alpha)=3\tan(\beta)$ We need to find general solution of $\beta$ for
$$(\sin(\alpha+\beta)+\cos(\alpha +2\beta)\sin(\beta))^2 = 4\cos(\alpha)\sin(\beta)\sin(\alpha+ \beta)$$
$$\tan(\alpha)=3\tan(\beta)$$
I took the first equation and managed to simplify it up till here:
$$(\cos(\beta)\sin(\alpha+2\beta))^2=4\cos(\alpha)\sin(\beta)\sin(\alpha+ \beta)$$
What should I do next? I tried to use $2\sin(\beta)\sin(\alpha+\beta)=\cos(\alpha)-\cos(\alpha+2\beta)$ in the RHS in the hopes of cancelling or doing something about the $\alpha+2\beta$ terms. But I cannot seem to boil it down further.
Thanks!

By request, here's a sketch of the solution I mentioned in a comment to the question. (There's probably a quicker path.)
I'll ignore the case $\beta=\pm\pi/2$.
Now, start by expanding the sides of OP's first equation, express them in terms of $p:=\tan\alpha$ and $q:=\tan\beta$, and then apply the substitution $p=3q$ from OP's second equation:
$$\begin{align}
\left(\sin(\alpha+\beta) + \cos(\alpha + 2\beta) \sin\beta\right)^2 &=
\left( 2\sin\alpha\cos^2\beta + 2 \cos\alpha \cos\beta \sin\beta-\sin\alpha \right)^2 \cos^2\beta \\[4pt]
&= \frac{(2\tan\alpha + 2 \tan\beta-\tan\alpha\sec^2\beta)^2}{\sec^4\beta} \cos^2\alpha\cos^2\beta\\[4pt]
&= \frac{(2p + 2q-p(1+q^2))^2}{(1+q^2)^2}\cos^2\alpha\cos^2\beta\\[4pt]
&= \frac{(p + 2q-pq^2)^2}{(1+q^2)^2}\cos^2\alpha\cos^2\beta \\[4pt]
&\overset{p=3q}{=}\;\frac{q^2(5-3q^2)^2}{(1+q^2)^2}\cos^2\alpha\cos^2\beta \tag1\\[4pt]
4 \cos\alpha \sin\beta \sin(\alpha+\beta) &=
4 \sin\beta (\sin\alpha\cos\beta + \cos\alpha \sin\beta)\cos\alpha \\[4pt]
&= 4\tan\beta\,(\tan\alpha +  \tan\beta) \cos^2\alpha \cos^2\beta \\[4pt]
&= 4q(p+q)\cos^2\alpha \cos^2\beta \\[4pt]
&\overset{p=3q}{=} 16q^2\cos^2\alpha \cos^2\beta \tag2
\end{align}$$
So, OP's first equation, $(1)=(2)$, becomes
$$\begin{align}
\frac{q^2(5-3q^2)^2}{(1+q^2)^2}\cos^2\alpha\cos^2\beta &\;=\; 16q^2\cos^2\alpha\cos^2\beta \\[4pt]
q^2(5-3q^2)^2 &\;=\; 16q^2(1+q^2)^2\\[6pt]
q^2 (7 q^2-1) (q^2+9)  &\;=\; 0 \tag3
\end{align}$$
Therefore, ignoring non-real values, we have $\tan\beta=0$ or $\tan\beta=\pm1/\sqrt{7}$; and, respectively, $\tan\alpha=0$ or $\tan\alpha=\pm3/\sqrt{7}$. $\square$
