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 !
 A: Hints:
$$2\sin3a=\sqrt2\iff \sin 3a=\frac{\sqrt2}2=\frac1{\sqrt2}\iff 3a=\begin{cases}\frac\pi4\\{}\\\frac{3\pi}4\end{cases}+2k\pi\;,\;k\in\Bbb Z\;\ldots$$
A: $2 \sin(3a)=\sqrt{2}$
$\implies \sin(3a)=\frac{\sqrt{2}}{2}$
$\implies \arcsin(\sin(3a))=\arcsin\left({\frac{\sqrt{2}}{2}}\right)$
$\implies 3a=\arcsin\left({\frac{\sqrt{2}}{2}}\right)$
Does that help?
A: Hint:
$$\sin x = y \iff x = \arcsin y + 2\,k\,\pi \;\;\mathrm{or}\;\; x = \pi - \arcsin y + 2\,k\,\pi $$
A:        2 sin (3a) = sqrt(2)
    =>    sin (3a) = sqrt(2) / 2 = 1/(sqrt2)  = sin (pi/4)

     =>        3a  = n pi + (-1)^n (pi/4) 

      =>         a  = n pi /3  +  (-1)^n (pi/12) for n belongs to integers 

A: 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. 
