Why can't a recurring pattern be found by using the second term in sine graph? In the question
$\sin(x) = \sin (4x), 0 < x < \pi$
I can use the equation
$4x = \pi - x$ ,
$4x = 2\pi + x$ ,
$4x = 3\pi - x$ ,
to work out the solutions, which are $\pi/5, 2\pi/3, 3\pi/5$.
But $\sin(x) = \sin(\pi - x)$.
I am thinking that multiple of the answer of second quadrant should also work.
So, if I try
$4(\pi - x) = 2\pi + x$ , it gives $2\pi/5$, and
$\sin\left(\dfrac{2\pi}{5}\right) = 0.951$
$\sin\left(4 \cdot \dfrac{2\pi}{5}\right) = -0.951$
No good.
If I try
$4(\pi - x) = 3\pi - x$ , it gives $\pi/4$, and
$\sin\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}$
$\sin\left(4 \cdot \frac{\pi}{4}\right) = 0$
Also, no good.
Can someone explain why does it not work please?
 A: The sine of an angle in standard position (vertex at the origin, initial side along the positive $x$-axis) is the $y$-coordinate of the point where the terminal side of the angle intersects the unit circle.

Clearly, $\sin\theta = \sin\varphi$ if $\theta = \varphi$.  By symmetry, $\sin\theta = \sin\varphi$ if $\theta = \pi - \varphi$.  Also, $\sin\theta = \sin\varphi$ if $\theta$ is coterminal with $\varphi$ or $\pi - \varphi$.  Hence, $\sin\theta = \sin\varphi$ if
$$\theta = \varphi + 2k\pi, k \in \mathbb{Z}$$
or
$$\theta = \varphi + 2m\pi, m \in \mathbb{Z}$$
At the risk of obscuring the symmetry argument above, $\sin\theta = \sin\varphi$ if
$$\theta = (-1)^n\varphi + n\pi, n \in \mathbb{Z}$$
You wish to solve the equation $\sin x = \sin(4x)$ in the interval $(0, \pi)$.  Let's apply the above formulas.
\begin{align*}
4x & = x + 2k\pi, k \in \mathbb{Z} & 4x & = \pi - x + 2m\pi, m \in \mathbb{Z}\\
3x & = 2k\pi, k \in \mathbb{Z} & 5x & = \pi + 2m\pi, m \in \mathbb{Z}\\
x & = \frac{2k\pi}{3}, k \in \mathbb{Z} & x & = \frac{\pi}{5} + \frac{2m\pi}{5}, m \in \mathbb{Z}
\end{align*}
which is the general solution of the equation $\sin(4x) = \sin x$.
In order to obtain values in the interval $(0, \pi)$, we must take $k = 1$ and can take $m = 0, 1$.  That yields the solutions
\begin{align*}
x & = \frac{2\pi}{3} & x & = \frac{\pi}{5}, \frac{3\pi}{5}
\end{align*}
which you found by applying the formula
$$\theta = (-1)^n\varphi + n\pi, n \in \mathbb{Z}$$
for particular values of $n$.
Let's apply the above formulas again, with the roles of $x$ and $4x$ reversed.
\begin{align*}
x & = 4x + 2j\pi, j \in \mathbb{Z} & x & = \pi - 4x + 2\ell \pi, \ell \in \mathbb{Z}\\
-3x & = 2j\pi, j \in \mathbb{Z} & 5x & = \pi + 2\ell \pi, \ell \in \mathbb{Z}\\
x & = -\frac{2j}{3}, j \in \mathbb{Z} & x & = \frac{\pi}{5} + \frac{2\ell\pi}{3}, \ell \in \mathbb{Z}
\end{align*}
If we set $k = -j$ and $m = \ell$, we obtain
\begin{align*}
x & = \frac{2k}{3}, k \in \mathbb{Z} & x & = \frac{\pi}{5} + \frac{2m\pi}{5}, m \in \mathbb{Z}
\end{align*}
which is the same general solution as we obtained above.  To find the particular solutions in the interval $(0, \pi)$, we proceed as above.
Notice I obtained the same solutions by reversing the roles of $x$ and $4x$, not $x$ and $\pi - x$.  While $\sin x = \sin(\pi - x)$, it does not follow that $\sin(4x) = \sin[4(\pi - x)] = \sin(4\pi - 4x)$, as your examples illustrate.  Observe that
$$\sin(4\pi - 4x) = \sin(4\pi)\cos(4x) - \cos(4\pi)\sin(4x) = -\sin(4x)$$
Therefore, by replacing $\sin(4x)$ with $\sin[4(\pi - x)]$, you replaced $\sin(4x)$ by $-\sin(4x)$, which is why you obtained
$$\sin\left(4 \cdot \frac{2\pi}{5}\right) = -\sin\left(\frac{2\pi}{5}\right)$$
You made an error when you solved the equation
$$4(\pi - x) = 3\pi - x$$
You should have obtained $x = \dfrac{\pi}{3}$.  Notice that
$$\sin\left(4 \cdot \frac{\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right)$$
which is also attributable to replacing $\sin(4x)$ by $\sin[4(\pi - x)] = -\sin(4x)$.
