Finding roots of a trigonometric function I have a calculus problem that has some trigonometric difficulty to it.
It is $\ 20\sin{x}-10\sin{2x}-\frac{40}{\pi}=0$. I basically want to find two $x \in [0,\pi]$.
I got to $\ \pi\sin{x}(1-\cos{x})=2$
I don't know if there is some trig identity or trick I am missing out on here but I am lost. I found the actual roots from my graphing calculator but would like to know how to go about doing this on my own.

Similar to Robert Israel's answer.
Using the tangent half-angle substitution $$t=\tan(\frac x 2)\qquad \sin(x)=\frac{2t}{1+t^2}\qquad \cos(x)=\frac{1-t^2}{1+t^2}$$ the equation reduces to $$t^4-2 \pi  t^3+2 t^2+1=0$$ Just as  Robert Israel answered, solving analytically quartic equations is not the most pleasant thing to do and numerical methods are required. 
Looking here, the discriminant is given by $\Delta=16 \pi ^2 \left(64-27 \pi ^2\right) <0$ and then the equation has two distinct real roots and two complex conjugate non-real roots.
Looking at the plot of the function, we can see that one root is "close" to $1$ and the second "close" to $6$. So, let us use Newton method to solve for the roots. The successive iterates will be 
$$\left(
\begin{array}{cc}
 n & t_n \\
 0 & 1.000000 \\
 1 & 0.789560 \\
 2 & 0.720530 \\
 3 & 0.712669 \\
 4 & 0.712570
\end{array}
\right)$$
$$\left(
\begin{array}{cc}
 n & t_n \\
 0 & 6.00000 \\
 1 & 5.94350 \\
 2 & 5.94182
\end{array}
\right)$$
which are the solutions for six significant figures.
So, $$t_1=0.712570 \implies \frac {x_1}2=\tan^{-1}(0.712570)\implies x_1=1.23822$$
 $$t_2=5.94182 \implies \frac {x_2}2=\tan^{-1}(5.94182)\implies x_2=2.80812$$
