Inequality for $\sin(20°)$ Prove that $$\frac{1}{3} < \sin{20°} < \frac{7}{20}$$
Attempt 
$$\sin60°=3\sin20°-4\sin^{3}(20°)$$
Taking $\sin20°$=x
I got the the equation as
$$8x^3-6x+\sqrt{3} =0$$ 
But from here I am not able to do anything. Any suggestions?
Thanks!
Edit-graph of p(x)

 A: Let $p(x)$ denote the cubic.
Since $$\lim_{x\to - \infty}p(x)=-\infty<0\quad \quad p(0)>0\quad \quad  p(.5)<0\quad \quad\lim_{x\to + \infty}p(x)=+\infty$$ we see that one root is negative, a second lies between $0$ and $.5$ and the third is greater than $.5$ 
Of course $\sin(20^{\circ})$ is positive and, since $0<20<30$ we see that $\sin(20^{\circ})<.5$  Thus the root we want is the middle one of the three.  We then remark that $$p\left( \frac 13 \right)>0 \quad \& \quad p\left( \frac 7{20}\right)<0$$  so the root we care about must be between $\frac 13$ and $\frac {7}{20}$ and we are done.
A: Mostly for fun, here's a way to show that
$$6\left(1\over3\right)-8\left(1\over3\right)^3\lt\sqrt3\lt6\left(7\over20\right)-8\left(7\over20\right)^3$$
with only a small amount of multi-digit arithmetic:
$$6\left(1\over3\right)-8\left(1\over3\right)^3=2-{8\over27}\lt2-{8\over28}=2-{2\over7}={12\over7}$$
and $12^2=144\lt147=3\cdot7^2$, which gives the first inequality, while
$$6\left(7\over20\right)-8\left(7\over20\right)^3={7\over10}\left(3-\left(7\over10\right)^2 \right)={7\over10}\left(300-49\over100 \right)\gt{7\over10}\cdot{250\over100}={7\over4}$$
and $7^2=49\gt48=3\cdot4^2$ gives the second inequality.
