I have solved it, but it does not match with the last part of solution. The logic is: Let's consider complex series $\sum\limits_{k=1}^{\infty} \frac{\cos 2k + i\sin 2k}{3^k}$, imaginary part of which is exactly the series in question.
The complex series is a complex geometric series $\sum\limits_{k=0}^{\infty} z^k$, having $z = \frac{\cos 2 + i\sin 2}{3}$ (De Moivre's formula).
The geometric series in quesion is decreasing and hence its sum equals to: $$\frac{1}{1-(\cos 2 + i\sin 2)/3}$$
The above is the sum including $k = 0$ which gives us $z= 1$. After subtracting it: $$\frac{1}{1-(\cos 2 + i\sin 2)/3} - 1 = \frac{1 - 1 + (\cos 2 + i\sin 2)/3}{1-(\cos 2 + i\sin 2)/3} = \frac{\cos2 + i \sin 2}{(3- \cos 2) - i \sin 2} = \mbox{multiply nominator and denominator so to get rid of imaginary part in denominator} = \frac{3\cos2 - \cos^2 2 + i\cos2\sin2 - i\cos2\sin2 - \sin^2 2 + i 3\sin2}{(3-\cos 2)^2 + \sin^2 2} = \frac{(3\cos2-1)+i3\sin2}{(3-\cos 2)^2 + \sin^2 2}$$
The imaginary part of the latter gives us the answer to the series in question: $$\sum\limits_{k=1}^{\infty} \frac{\sin 2k}{3^k} = \frac{3\sin2}{10 - 6\cos 2}.$$
In the solution the last part is somehow simplified to $\dfrac{\sin 2}{4 - 2\cos 2}$.
Am I missing something or is it just some typo in the final answer?