Solve $\sin 5x = \sin x$. I have to solve the following equation:
$$\sin 5x = \sin x$$
This is what I tried:
$$5x = x + 2k \pi \hspace{5cm} 5x = \pi - x + 2k \pi$$
$$4x = 2k \pi \hspace{5cm} 6x = (2k + 1) \pi$$
$$\hspace{2cm} x =  k \dfrac{\pi}{2}, k\in \mathbb{Z} \hspace{4cm} x = (2k + 1)\dfrac{\pi}{6}, k \in \mathbb{Z} \hspace{2cm}$$
So we have that:
$$x \in \bigg \{ k \dfrac{\pi}{2} | k \in \mathbb{Z} \bigg \} \cup \bigg \{ (2k+1) \dfrac{\pi}{6} | k \in \mathbb{Z} \bigg \}$$
The problem is that my textbook has the following answers listed:
A. $ \bigg \{ \dfrac{k \pi}{ 5 - (-1)^k } | k \in \mathbb{Z} \bigg \}$
B. $\bigg \{ \dfrac{k\pi}{5} | k \in \mathbb{Z} \bigg\}$
C. $\bigg \{ \dfrac{k\pi}{10} | k \in \mathbb{Z} \bigg \}$
D. $\bigg \{ (-1)^k \arcsin \dfrac{1}{5} + k\pi | k \in \mathbb{Z} \bigg \}$
E. $\bigg \{ (-1)^k \dfrac{\pi}{3} + k\pi | k \in \mathbb{Z} \bigg \}$
None of the listed answers look even remotely like my own. So did I do something wrong or is it one of those cases where the same answer can be written in multiple ways?
 A: Hint:
$$k\frac{\pi}{2} = \frac{2k\pi}{4}= \frac{2k\pi}{5-(-1)^{2k}}$$
$$(2k+1)\frac{\pi}{6} = \frac{(2k+1)\pi}{5-(-1)^{2k+1}}$$
A: 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.
A: Using a bit of algebra.
Expand $\sin(5x)$ to get
$$\sin(5x)=\sin ^5(x)+5 \sin (x) \cos ^4(x)-10 \sin ^3(x) \cos ^2(x)$$ Use $\cos^2(x)=1-\sin^2(x)$ to get with $s=\sin(x)$
$$16 s^5-20 s^3+5 s=s \implies 16 s^5-20 s^3+4 s=0\implies 4s(4s^4-5s^2+1)=0$$ The expression in brackets is a quadratic in $s^2$ and then the solutions of the proposed equation are
$$s=0 \qquad s=\pm \frac 12\qquad s=\pm 1$$
These correspond to angles you know. Now, use the periodicity of $\sin(x)$ and then your very correct result.
