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.
 A: If $t = \sin(x)$, this can be written as $\pi t (1\pm\sqrt{1-t^2}) = 2$,
or $\pm\sqrt{1-t^2} = 2/(\pi t) - 1$.  Squaring both sides and simplifying, we get the quartic
$$ \pi^2 t^4-4 \pi t+4 = 0$$
which has two real roots, approximately $.3273258251$ and $.9452058404$.
There are formulas for solving the quartic in radicals, but they aren't 
terribly pretty.
For $t = .3273258251$ we want $\cos(x) < 0$, thus $x = \pi - \arcsin(t) \approx 2.808120551$, while for $t = .9452058404$
we want $\cos(x) > 0$, thus $x = \arcsin(.9452058404) \approx 1.238224522$.
A: 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, $$t_1=0.712570 \implies \frac {x_1}2=\tan^{-1}(0.712570)\implies x_1=1.23822$$
 $$t_2=5.94182 \implies \frac {x_2}2=\tan^{-1}(5.94182)\implies x_2=2.80812$$
