What this sine function equation means?

Question

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 $\sin\frac{\pi}{6}=\frac{1}{2}$?

Answer 1

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.

Answer 2

We simply need to throw out the "solution" $y = \sin\left(\frac \pi 6\right) = -1\,$ because for $0 \lt \theta \lt \pi$, $0 \lt \sin\theta \lt 1$.