Solving a trigonometric equation: $2 \sin(3a)=\sqrt{2}$ I have the following equation :

$2 \sin(3a)=\sqrt{2}$

Not sure how to solve it (Because it's a transformed sin function, meaning 6 solution with 3 cycles in $2\pi$) after a moment I finally found that each solution can be expressed as a multiple of $\frac{\pi}{12}$ (I aided myself with a graph of the function.)

For example :

$\frac{\pi}{12} \frac{3\pi}{12} \frac{9\pi}{12}$  etc. 

My question would be : Is this the standard way of solving transformed functions of this sort ? Is there a sort of symbolic way without the need to use the graph ?

Is my way complicated for nothing ? Thank you !
Hints:

$$2\sin3a=\sqrt2\iff \sin 3a=\frac{\sqrt2}2=\frac1{\sqrt2}\iff 3a=\begin{cases}\frac\pi4\\{}\\\frac{3\pi}4\end{cases}+2k\pi\;,\;k\in\Bbb Z\;\ldots$$
We have $2sin(3A)=\sqrt 2$.

Multiplying both sides by $\sqrt 2$ , we have

$2 \sqrt 2sin(3A)=2$

$\implies sin(3A)=\frac {1}{\sqrt2}$. Let $3A=\alpha$.Then

$sin(\alpha)=\frac {1}{\sqrt2}$.Clearly such a value for $\alpha$ which satisfies the equation is $\alpha=30$.But $\alpha = 3A$ $\implies 3A =30$.$\implies A=10$.Hope it helps.Any doubts are welcomed.