<s>[INST] Solve $\sin x = 1 - x$ How would you be solve sin x = 1 - x, without drawing the graph and manually measuring the point of intersection?
 [/INST] As @Gerben answered, building the Taylor series around $x=\frac \pi 6$ gives
$$y=x+\sin(x)-1=\frac{\pi -3}{6} +\frac{2+\sqrt{3}}{2} x+\sum_{n=2}^\infty \frac{\sqrt{3} \sin \left(\frac{\pi  n}{2}\right)+\cos \left(\frac{\pi n}{2}\right)}{2 n!} \left(x-\frac{\pi }{6}\right)^n$$
Truncated to $O\left(\left(x-\frac{\pi }{6}\right)^6\right)$ for example and using series reversion, then
$$x=\frac{\pi }{6}+t+\left(1-\frac{\sqrt{3}}{2}\right)
   t^2+\left(3-\frac{5}{\sqrt{3}}\right) t^3+\left(\frac{67}{6}-\frac{77}{4
   \sqrt{3}}\right) t^4+\left(\frac{232}{5}-\frac{803}{10 \sqrt{3}}\right)
   t^5+O\left(t^{6}\right)$$ where $t=\frac{y+\frac{3-\pi }{6}}{1+\frac{\sqrt{3}}{2}}$.
Making $y=0$ that is to say $t=-\frac{\left(2-\sqrt{3}\right) (\pi -3)}{3} $, this gives
$x=0.51097342938847$ while the exact solution is
$x=0.51097342938857$.
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