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!
 A: 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 $\tan\alpha=\pm3/\sqrt{7}$. $\square$
A: You have attacked the wrong equation. You should have started with the second equation as shown below. It can be written as,
$$\sin\left(\alpha\right) \cos\left(\beta\right)=3\sin\left(\beta\right) \cos\left(\alpha\right).$$
Now we add both sides $\sin\left(\beta\right) \cos\left(\alpha\right)$ to get,
$$\sin\left(\alpha\right) \cos\left(\beta\right)+\sin\left(\beta\right) \cos\left(\alpha\right)=4\sin\left(\beta\right) \cos\left(\alpha\right),$$
which can be simplified to obtain,
$$\sin\left(\alpha+\beta\right)=4\sin\left(\beta\right) \cos\left(\alpha\right).$$
Now, we use this to simplify the first equation.
$$\left(\sin\left(\alpha+\beta\right)+ \cos\left(\alpha+2\beta\right) \sin\left(\beta\right)\right)^2=\sin^2\left(\alpha+\beta\right)$$
This yields two equations, i.e.,
$$\cos\left(\alpha+2\beta\right) \sin\left(\beta\right)=0\qquad\qquad\text{and}\tag{1}$$
$$\cos\left(\alpha+2\beta\right) \sin\left(\beta\right)=-2\sin\left(\alpha+\beta\right).\tag{2}$$
Let us consider (1). It gives two solutions, i.e., $\space\sin\left(\beta\right)=0\space$ and $\space\cos\left(\alpha+2\beta\right)=0.\space$ The first of these two gives us,
$$\sin\left(\beta\right)=0\quad\rightarrow\quad \beta=0.$$
This is one of the solutions mentioned in the comment by @Blue. To obtain the value of $\beta$ from the other one needs some doing.
$$\cos\left(\alpha+2\beta\right)=0 \quad\rightarrow\quad \alpha=\dfrac{\pi}{2}-2\beta$$
We feed this into the left-hand side of the second of the two equations given in OP’s problem statement.
$$\small{\tan\left(\alpha\right)=\tan\left(\dfrac{\pi}{2}-2\beta\right)=\dfrac{1}{\tan\left(2\beta\right)}=\dfrac{1-\tan^2\left(\beta\right)}{2\tan\left(\beta\right)}=3\tan\left(\beta\right)\enspace\rightarrow\space 7\tan^2\left(\beta\right)=1\enspace\rightarrow\space \tan\left(\beta\right)=\pm\dfrac{1}{\sqrt{7}}}$$
This agrees with the @Blue’s second set of solutions. However, the story does not ends here. You have to find out whether (2) also has solutions other than the ones mentioned here.
