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?
Best Answer
Hint:
$$k\frac{\pi}{2} = \frac{2k\pi}{4}= \frac{2k\pi}{5-(-1)^{2k}}$$
$$(2k+1)\frac{\pi}{6} = \frac{(2k+1)\pi}{5-(-1)^{2k+1}}$$