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!

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.
