How to solve $x-\sin x = \dfrac {\pi}{10}$? Solve $x-\sin x = \dfrac {\pi}{10}$.
I got $1.26$ using Newton Raphson method but Is there any other alternative method?
 A: You could try with fixed point iteration, where you have two curves $f(x)=x$ and $g(x)=\sin{x}+\frac{\pi}{10}$ and you compute the intersection of both with a guessed $x$ value, being the seed $x^0$ .
This scheme is presented as follows:
$$x^{n+1}=g(x^{n})=\sin{x^n}+\frac{\pi}{10} \tag{1}$$
This scheme converges to the root $x^*$  within an interval in which the derivative of $g(x)$ is less that unity, $i.e$ $x^0\in(-\pi,\pi)$.
This can be proved computing the error:
$$|x^{n+1}-x^*|=|g(x^n)-g(x^*)|\leq L|x^n-x^*|$$
The bounding constant $L$ must be less that unity for $(1)$ to converge. This $L$ coincides with:
$$L=\max{|g'(x)|}$$ 
Since $g'(x) = \cos{x}$ and its maximum value is 1 in $x=\pm \pi$, equation $(1)$ will always converge to $x^*$ if the seed is chosen within $x^0\in(-\pi,\pi)$
A: One other method is to use fixed point iteration. Rewrite: $x = \phi(x)=\sin(x)+\pi/10$. The iteration is
$x_{k+1}=\phi(x_{k})=\sin(x_k)+\pi/10$.
Start with $x = 0$ and iterate 8-9 steps and you will get a pretty good numerical solution.
A: defining $$f(x)=x-\sin(x)-\frac{\pi}{10}$$ and $$f'(x)=1-\cos(x)\geq 0$$ and $$f(0)<0$$ and $$f(2\pi)>0$$ ntherefore we get only one solution $$x\approx 1.26894786483261674188$$ 
