<s>[INST] 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 $\sin\frac{\pi}{6}=\frac{1}{2}$?
 [/INST] We simply need to throw out the "solution" $y = \sin\left(\frac \pi 6\right) = -1\,$ because for $0 \lt \theta \lt \pi$, $0 \lt \sin\theta \lt 1$.
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