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)


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.
