Solve $\sin 5x = \sin x$.

Hint:

$$k\frac{\pi}{2} = \frac{2k\pi}{4}= \frac{2k\pi}{5-(-1)^{2k}}$$

$$(2k+1)\frac{\pi}{6} = \frac{(2k+1)\pi}{5-(-1)^{2k+1}}$$

Let's get rid of the confusing notation and actually look at the permissible values of $x$. The values $k\pi/2$ correspond to $\ldots,-\pi,-\pi/2,0,\pi/2,\pi,\ldots$ whereas the values $(2k+1)\pi/6$ correspond to $\ldots-\pi/6,\pi/6,\pi/2,5\pi/6,7\pi/6,3\pi/2,\ldots$.

So now the question is which of the answers this is equivalent to. There's certainly no $\arcsin(1/5)$ involved here, so option D is out immediately. With $k=1$, option C gives $x=\pi/10$ which is not in the solution set, so this is out too. Similarly $k=1$ with option B gives $x=\pi/5$, so that is out too. For option E, if $k=1$ we get $2\pi/3=4\pi/6$, which is again not in the solution set.

Now if you list out all the values of option A you will actually see that it is, in fact, equivalent to what you got, so A is the answer. To show this rigorously, you can use the hint provided in Flame Trap's answer.