How to find the general solution of $\sin \left(\theta +\frac{\pi }{6}\right) =\cos 3\theta $ Find the general solution of the equation.

\begin{eqnarray} \sin \left(\theta +\frac{\pi }{6}\right)\space
 =\space \cos 3\theta \\ \end{eqnarray}

The answers in my book are $\left(6n+1\right)\frac{\pi }{12},\space \left(n-\frac{1}{6}\right)\pi $.
Should I first convert all the things to cos(xxx)=??? or sin(xxx)=???.
So I try :

\begin{eqnarray}
\sin \theta \cos \frac{\pi }{6}+\cos \theta \sin \frac{\pi }{6}&=&\cos 2\theta \cos \theta -\sin 2\theta \sin \theta \\
\sin \theta \cos \frac{\pi }{6}+\sin 2\theta \sin \theta &=&\cos 2\theta \cos \theta -\cos \theta \sin \frac{\pi }{6}\\
\sin \theta \left(\sin 2\theta +\cos \frac{\pi }{6}\right)&=&\cos \theta \left(\cos 2\theta -\sin \frac{\pi }{6}\right)\\
\\?...&=&...?\\
\end{eqnarray}

What should I do?
I do not know whether my direction is correct.
Thank you for your attention.
Update 1 :

I can get the first ans by using method 1 but I cannot find the next one$\space \left(n-\frac{1}{6}\right)\pi $.

Method $1:$
As $\cos(\frac\pi2-x)=\sin x$
$$\cos3\theta=\sin\left(\theta+\frac\pi6\right)=\cos\{\frac\pi2-\left(\theta+\frac\pi6\right)\}$$
$$\cos3\theta=\cos \left(\frac\pi3-\theta\right)$$
Now, we know $\cos x=\cos A\implies x=2n\pi\pm A$ where $n$ is any integer
Method $2:$
As $\sin(\frac\pi2-x)=\cos x$
$$\sin\left(\theta+\frac\pi6\right)=\cos3\theta=\sin\left(\frac\pi2-3\theta\right)$$
Now, we know $\sin x=\sin B\implies x=m\pi+(-1)^m B$ where $m$ is any integer
