Solving $\sin(5x) = \sin(x)$ If I have an equation:
$$\sin(5x) = \sin(x)$$
In what case can I equate $$5x = x$$
Is it only when there is a multiply of $2\pi n$ on either side, where n is any integer so
$$ 5x = x+2\pi n$$
Also with this method can I get every possible solution or does that not work?
I know I can use the sine addition formula but I want to see others I way I can solve this.
 A: You might find this useful:
$$
\sin(A)=\sin(B) \iff A=B +n2\pi\ \mbox{ or }\ A=\pi-B +n2\pi,
$$
with $n\in\mathbb{Z}$. 
$$
\cos(A)=\cos(B) \iff A=B +n2\pi\ \mbox{ or }\ A=-B +n2\pi,
$$
with $n\in\mathbb{Z}$. 
So you need to solve two equations separately:
$$5x=x+n2\pi$$
and 
$$5x=\pi-x+n2\pi.$$
Good luck :)

Edit:
One way to prove 
$$
\sin(x)=\sin(y) \iff x=y +n2\pi\ \mbox{ or }\ x=\pi-y +n2\pi,
$$
is to just verify the solutions work, using a consequence of Euler's formula:
$$\sin(x)=\frac{\mathrm{e}^{ix}-\mathrm{e}^{-ix}}{2i}.$$
So, here we go:
$$
\sin(x)=\sin(y) \iff \frac{\mathrm{e}^{ix}-\mathrm{e}^{-ix}}{2i}=\frac{\mathrm{e}^{iy}-\mathrm{e}^{-iy}}{2i} \iff \mathrm{e}^{ix}-\mathrm{e}^{-ix}=\mathrm{e}^{iy}-\mathrm{e}^{-iy}.
$$


*

*Now note that $x=y+n2\pi$ is indeed a solution:
$$\mathrm{e}^{ix}-\mathrm{e}^{-ix}=\mathrm{e}^{i(y+n2\pi)}-\mathrm{e}^{-i(y+n2\pi)}=\mathrm{e}^{iy}\mathrm{e}^{in2\pi}-\mathrm{e}^{-iy}\mathrm{e}^{-in2\pi}=\mathrm{e}^{iy}-\mathrm{e}^{-iy}
$$

*and that $x=\pi-y+n2\pi$ is indeed a solution:
$$\mathrm{e}^{ix}-\mathrm{e}^{-ix}=\mathrm{e}^{i(\pi-y+n2\pi)}-\mathrm{e}^{-i(\pi-y+n2\pi)}=\mathrm{e}^{-iy}\mathrm{e}^{i(2n+1)\pi}-\mathrm{e}^{iy}\mathrm{e}^{-i(2n+1)\pi}=\mathrm{e}^{iy}-\mathrm{e}^{-iy}.$$


The last equalities use $\mathrm{e}^{in\pi}$ is equal to $1$ if $n$ is even and equal to $-1$ if $n$ is odd.
A: Suppose $\sin x=\sin y$. Then for the cosine there are just two possibilities, since $\cos^2x=1-\sin^2x=1-\sin^2y=\cos^2y$: either $\cos x=\cos y$ or $\cos x=-\cos y$.
In the first case, the angles $x$ and $y$ must differ by an integer multiple of $2\pi$; in the second case, if we write $z=\pi-y$, we have $\sin z=\sin y$ and $\cos z=-\cos y$, so $\sin x=\sin z$ and $\cos x=\cos z$, so we're in the above case and we conclude $x$ differs from $z=\pi-y$ by an integer multiple of $2\pi$.
Therefore the equation $\sin x=\sin y$ is solved by the two cases


*

*$x=y+2k\pi$

*$x=\pi-y+2k\pi$


In your particular case the first one becomes $5x=x+2k\pi$ that can also be written
$$
x=k\frac{\pi}{2}
$$
and the second case becomes $5x=\pi-x+2k\pi$, that is,
$$
x=\frac{\pi}{6}+k\frac{\pi}{3}
$$
Some solutions, for instance $\pi/2$ appear in both sets. 

With a similar reasoning you can solve $\cos x=\cos y$ with the two sets


*

*$x=y+2k\pi$

*$x=-y+2k\pi$


If you have $\sin x=\cos y$, you reduce it to the previous cases by noting it is the same as $\sin x=\sin(\pi/2-y)$ (or $\cos(\pi/2-x)=\cos y$).
A: You may also use this identity:
\begin{align}
\sin(5x)-\sin(x)&=2\sin(\frac{5x-x}{2})\cos(\frac{5x+x}{2})\\
&=2\sin(2x)\cos(3x)\\
\end{align}
A: $$\sin5x=\sin x\Rightarrow 5x=x+n.2\pi$$ or $$5x=\pi-x+n.2\pi$$ 
which leads to solutions of the form $$x=n.\frac{\pi}{2}\cup x=(2n+1)\frac{\pi}{6}$$
A: Apart from trivial solution ($x=y+2n\pi$), you can see others geometrically from looking at the Unit circle:

For angle $x$ in $1$st quadrant you have $y$ from $2$nd quadrant such $y = \pi - x$ which gives you $\sin x =\sin y$. Taking periodicity into account, you have
$$
y = \pi - x + 2n\pi
$$
Similarly for $x$ being from $3$rd quadrant you see that it will have same $\sin$ value as $y$ from $4$th quadrant satisfying $2\pi - y = x - \pi$, which is just the same equality as above in disguise (put $n=1$).
So overall you have
$$
y = x + 2n\pi
$$
$$
y = \pi - x + 2n\pi
$$
for $n \in \mathbb{Z}$. Now just plug in your $x$ and $5x$ for $x$ and $y$ to get your solution.
A: Alternative solution: expand the LHS:
$$16\sin^5 x - 20\sin^3 x + 5\sin x = \sin(5x) = \sin x,$$
and $s = \sin x$ is a solution of
$$(16s^4 - 20s^2 + 4)s = 16s^5 - 20s^3 + 4s = 0.$$
