What this sine function equation means? Apostol's book "Calculus" asks to prove that
$$\sin\frac{\pi }{6}=\frac{1}{2}$$
using the fact that 
$$\sin 3x=3\sin x-4\sin^3 x$$
and
$$\sin \frac{\pi}{2}=1$$
So, we take $x=\frac{\pi}{6}$ and we have
$$\sin\frac{\pi}{2}=3\sin\frac{\pi}{6}-4\sin^3 \frac{\pi}{6}$$
$$1=3\sin\frac{\pi}{6}-4\sin^3 \frac{\pi}{6}$$
if we take $y=\sin \frac{\pi}{6}$ 
$$4y^3-3y+1=0$$
$$\left( {y- \frac{1}{2}} \right)^2(y+1)=0$$
and finally $y=\sin\frac{\pi}{6}=\frac{1}{2}$ or $y=\sin\frac{\pi}{6}=-1$. What it means?, it not should only be $\sin\frac{\pi}{6}=\frac{1}{2}$?

If it is true that $\sin\frac\pi6=\frac12$, then it is true that EITHER $\sin\frac\pi6=\frac12$ OR $\sin\frac\pi6=-1$.
There are two different angles that, when multiplied by $3$, give angles whose sine is $1$.  (I am considering two angles the same if they differ by a multiple of $2\pi$.)  Those two angles are $\pi/6$ and $-\pi/2$.  Three times a $30^\circ$ turn to the left is a $90^\circ$ turn to the left, and three times a $90^\circ$ turn to the right is also a $90^\circ$ turn to the left.  That's why you get two solutions.
