How do I find the angle $\theta$ This is the equation:
$$\sin\theta=0.8\theta$$ 
 A: Clearly there are three solutions:

Clearly $\theta = 0$ is a solution.  By the antisymmetry of the component functions we know there are two symmetric solutions.  Simple numerical solution by computer gives the other two:  $\theta = \pm 1.13.$
A: Using Taylor's series up to the third term one gets the approximate polynomial equation:
$$\theta-\frac{\theta^3}{6}+\frac{\theta^5}{120}=\frac{4}{5}\theta$$
from which we obtain
$$\theta\cdot \left(24-20\cdot \theta^2+\theta^4\right)=0$$
which has the solution $\theta=0$ and the two solutions to the biquadratic equation
$$\theta^2=\frac{20\pm\sqrt{304}}{2}$$
from which we get 
$$\theta=\pm\sqrt{\frac{20-\sqrt{304}}{2}}\cong \pm 1.132343637293314$$
Comment: I included only the first three terms in order to get a biquadratic. Otherwise I would have got a polynomial equation of degree 3, which is harder to deal with. Moreover, the approximation works quite well because we know (a priori) that the solution is not too big, as it can be at most $1/0.8=1.25$.
