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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 !

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What happens when you take $\arcsin$ of both sides? – Igor Rivin Dec 6 '13 at 22:50
Well when I take the arcsin of sqrt(2)/2 I get pi/4, but my calculator considers it like a normal sin function without any transformation – Astroman Dec 6 '13 at 22:54

5 Answers 5


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

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why did you put it in the form 1/sqrt(2) – Astroman Dec 6 '13 at 23:19
Habit, @Astroman: it's the same, of course, and it's one of the basic values of basic angles for sine and cosine. – DonAntonio Dec 6 '13 at 23:23
This is the best answer +1 – user66360 Dec 6 '13 at 23:54

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

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

$\implies \arcsin(\sin(3a))=\arcsin\left({\frac{\sqrt{2}}{2}}\right)$

$\implies 3a=\arcsin\left({\frac{\sqrt{2}}{2}}\right)$

Does that help?

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Well I did find pi/12 (Which I got in a geometrical way) – Astroman Dec 6 '13 at 22:59
First of all, let's assume $0\le a \le 2\pi$. Then I'd just look at the unit circle. You see two places in which $y=\frac{\sqrt{2}}{2}$: namely at $\frac{\pi}{4}$ and $\frac{3\pi}{4}.$ Hence, $\arcsin\left({\frac{\sqrt{2}}{2}}\right)=$$\frac{\pi}{4}$ and $\frac{3\pi}{4}$. If you divide by 3, you have $\frac{\pi}{12}$ and $\frac{3\pi}{12}.$ – user66360 Dec 6 '13 at 23:05
ok.... and after ? – Astroman Dec 6 '13 at 23:18
$a=\frac{\pi}{12}$ and $\frac{3\pi}{12}$. Was the explanation unclear? – user66360 Dec 6 '13 at 23:21
No, I understand but I'm not sure how to have the other 4 solutions ... – Astroman Dec 6 '13 at 23:22


$$\sin x = y \iff x = \arcsin y + 2\,k\,\pi \;\;\mathrm{or}\;\; x = \pi - \arcsin y + 2\,k\,\pi $$

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       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 
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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.

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