<s>[INST] How to solve $x= \sin(k- x)$? Is there a way to solve $x = \sin (k-x)$ without a computer, that is with a pocket calculator or pencil and paper?
 [/INST] Let $f(x) = x - \sin(k-x)$. It is trivial that there is no zero when $|x| > 1$. We also know that $f(0) = -\sin(k) \ne 0$.
The problem with the Newton's method is that it will fail when $f'(x) = 0$.
Let $x-k=y$, $g(y) = f(y+k) = y + k + sin(y)$. Let's also constrain $k \in [-pi/2, pi/2]$.
Using the Taylor's series for $sin(y)$, $g(y) \approx y + k + y - y^3/6$.
Thus:
$$
g(y) = 0 \\
k + 2y - y^3/6 = 0 \\
y^3/6 - 2y - k = 0 \\
y^3 - 12y - 6k = 0
$$
I used the WolframAlpha to help solving this polinomial.
All three solutions can be seen here
You can notice that only one solution is between $[-1, 1]$. Take a look at this solution.
Let $w = 3k + \sqrt{-64+9 k^2} = 3k + i\sqrt{64-9 k^2}$. But, as $|w| = 8$, it can be rewritten as $w = 8 cis(\alpha)$, $\alpha = arg(w) = acos(3k/8)$.
Let $z = \sqrt[3]{w} = 2cis(\alpha/3)$.
Thus, our root is $y = \frac{2i(\sqrt{3}+i)}{z} - \frac{(1+i\sqrt{3})z}{2}$. Using the polar form for complex numbers, $y = 2cis\left(\frac{2pi}{3}-\frac{\alpha}{3}\right) - 2cis\left( \frac{\pi}{3} + \frac{\alpha}{3} \right)$.
But we already know that $y$ is a real number. So, let's get only the real part.
$$
y = 2cis\left(\frac{2pi}{3}-\frac{\alpha}{3}\right) - 2cis\left( \frac{\pi}{3} + \frac{\alpha}{3} \right) \\
y = 2\cos\left(\frac{2pi}{3}-\frac{\alpha}{3}\right) - 2\cos\left( \frac{\pi}{3} + \frac{\alpha}{3} \right) \\
y = -4 \sin\left( \frac{\pi}{6} - \frac{\alpha}{3} \right) \\
y = -4 \sin\left( \frac{\pi}{6} - \frac{cos^{-1}(3k/8)}{3} \right)
$$
Finally, $x = k -4 \sin\left( \frac{\pi}{6} - \frac{cos^{-1}(3k/8)}{3} \right)$, if $k \in [-\pi/2, \pi/2]$.
Let's check: $$
k=\pi/4, x = 0.38744, f(x)=-0.00009688 \\
k=\pi/3, x = 0.51073, f(x)=-0.00037297
$$
Notice that any value of $k$ can be mapped in the interval $[-\pi, \pi]$.
We still have to solve for $k \in [-pi, -pi/2)$ and $k \in (pi/2, \pi]$. But, notice that, for k in these intervals, we can just map $k$ to the interval $[-\pi/2, \pi/2]$ and invert the sign of $sin(x-k)$.
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